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Ray Optics and Optical Instruments
Physics XII - Chapter 09: Complete NCERT Exercise Solutions
Master Ray Optics with NCERT solutions for reflection, refraction, lenses, mirrors, and optical instruments.
Question 9.1
A small candle, 2.5 cm in size, is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer & Explanation:
Step 1: Find focal length
For a concave mirror, \( R = -36 \, \text{cm} \) (negative by sign convention).
\( f = \frac{R}{2} = \frac{-36}{2} = -18 \, \text{cm} \).
Step 2: Apply mirror formula
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
Given \( u = -27 \, \text{cm} \), \( f = -18 \, \text{cm} \).
\( \frac{1}{-18} = \frac{1}{v} + \frac{1}{-27} \).
\( \frac{1}{v} = \frac{1}{-18} + \frac{1}{27} \).
Solve for \( v \) to get image distance.
Step 3: Find image size and nature
Use magnification \( m = -\frac{v}{u} = \frac{h_i}{h_o} \).
Calculate \( h_i \).
Step 4: Effect of moving candle closer
As object moves towards mirror (u decreases in magnitude), v increases. Screen must be moved away from mirror to capture sharp image.
For a concave mirror, \( R = -36 \, \text{cm} \) (negative by sign convention).
\( f = \frac{R}{2} = \frac{-36}{2} = -18 \, \text{cm} \).
Step 2: Apply mirror formula
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
Given \( u = -27 \, \text{cm} \), \( f = -18 \, \text{cm} \).
\( \frac{1}{-18} = \frac{1}{v} + \frac{1}{-27} \).
\( \frac{1}{v} = \frac{1}{-18} + \frac{1}{27} \).
Solve for \( v \) to get image distance.
Step 3: Find image size and nature
Use magnification \( m = -\frac{v}{u} = \frac{h_i}{h_o} \).
Calculate \( h_i \).
Step 4: Effect of moving candle closer
As object moves towards mirror (u decreases in magnitude), v increases. Screen must be moved away from mirror to capture sharp image.
Question 9.2
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer & Explanation:
Step 1: Sign conventions and formula
For convex mirror: \( f = +15 \, \text{cm} \), \( u = -12 \, \text{cm} \).
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
Step 2: Calculate image distance (v)
\( \frac{1}{15} = \frac{1}{v} + \frac{1}{-12} \).
\( \frac{1}{v} = \frac{1}{15} + \frac{1}{12} \).
Solve for \( v \). Value will be positive, indicating virtual image behind mirror.
Step 3: Find magnification (m)
\( m = -\frac{v}{u} \). Calculate m. It will be positive and less than 1, indicating erect, diminished image.
Step 4: Needle moved farther away
As \( |u| \) increases, \( |v| \) approaches \( f \) from the mirror side, image becomes smaller and remains virtual and erect.
For convex mirror: \( f = +15 \, \text{cm} \), \( u = -12 \, \text{cm} \).
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
Step 2: Calculate image distance (v)
\( \frac{1}{15} = \frac{1}{v} + \frac{1}{-12} \).
\( \frac{1}{v} = \frac{1}{15} + \frac{1}{12} \).
Solve for \( v \). Value will be positive, indicating virtual image behind mirror.
Step 3: Find magnification (m)
\( m = -\frac{v}{u} \). Calculate m. It will be positive and less than 1, indicating erect, diminished image.
Step 4: Needle moved farther away
As \( |u| \) increases, \( |v| \) approaches \( f \) from the mirror side, image becomes smaller and remains virtual and erect.
Question 9.3
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer & Explanation:
Step 1: Find refractive index of water (nw)
Real depth = 12.5 cm, Apparent depth = 9.4 cm.
\( n_{\text{water}} = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{12.5}{9.4} \).
Calculate value.
Step 2: Apparent depth with new liquid
For new liquid, \( n_l = 1.63 \).
Apparent depth' = \( \frac{\text{Real depth}}{n_l} = \frac{12.5}{1.63} \).
Step 3: Distance microscope must be moved
Microscope initially focused at 9.4 cm (for water).
New focusing distance = Apparent depth'.
Shift = New apparent depth - Old apparent depth.
Since new apparent depth is smaller, microscope must be moved downward.
Real depth = 12.5 cm, Apparent depth = 9.4 cm.
\( n_{\text{water}} = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{12.5}{9.4} \).
Calculate value.
Step 2: Apparent depth with new liquid
For new liquid, \( n_l = 1.63 \).
Apparent depth' = \( \frac{\text{Real depth}}{n_l} = \frac{12.5}{1.63} \).
Step 3: Distance microscope must be moved
Microscope initially focused at 9.4 cm (for water).
New focusing distance = Apparent depth'.
Shift = New apparent depth - Old apparent depth.
Since new apparent depth is smaller, microscope must be moved downward.
Question 9.4
Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface. [Given: Refractive index of glass, \( n_g = 1.52 \), Refractive index of water, \( n_w = 1.33 \)]
Answer & Explanation:
Step 1: Understand the data from figures
From Fig (a): Air to glass, \( i = 60^\circ \), find \( r \) using \( n_g = 1.52 \).
Snell's Law: \( 1 \times \sin 60^\circ = 1.52 \times \sin r_a \).
Step 2: Find refractive index relationship
From Fig (b): Air to water, \( i = 60^\circ \), \( n_w = 1.33 \).
\( 1 \times \sin 60^\circ = 1.33 \times \sin r_b \).
Step 3: Solve for water-glass interface
Case: Water to glass, \( i = 45^\circ \).
\( n_w \sin i = n_g \sin r \).
\( 1.33 \times \sin 45^\circ = 1.52 \times \sin r \).
Solve for \( r \), the angle of refraction in glass.
From Fig (a): Air to glass, \( i = 60^\circ \), find \( r \) using \( n_g = 1.52 \).
Snell's Law: \( 1 \times \sin 60^\circ = 1.52 \times \sin r_a \).
Step 2: Find refractive index relationship
From Fig (b): Air to water, \( i = 60^\circ \), \( n_w = 1.33 \).
\( 1 \times \sin 60^\circ = 1.33 \times \sin r_b \).
Step 3: Solve for water-glass interface
Case: Water to glass, \( i = 45^\circ \).
\( n_w \sin i = n_g \sin r \).
\( 1.33 \times \sin 45^\circ = 1.52 \times \sin r \).
Solve for \( r \), the angle of refraction in glass.
Question 9.5
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer & Explanation:
Step 1: Concept of critical angle
Light emerges only from a circular area where angle of incidence at water-air interface is less than critical angle \( i_c \).
\( \sin i_c = \frac{1}{n_w} = \frac{1}{1.33} \).
Step 2: Relate to geometry
Depth of bulb, \( h = 80 \, \text{cm} \).
Radius of circular area, \( r = h \tan i_c \).
Find \( i_c \) from \( \sin i_c \), then find \( \tan i_c \).
Step 3: Calculate area
Area = \( \pi r^2 = \pi (h \tan i_c)^2 \).
Substitute values to get area in cm² or m².
Light emerges only from a circular area where angle of incidence at water-air interface is less than critical angle \( i_c \).
\( \sin i_c = \frac{1}{n_w} = \frac{1}{1.33} \).
Step 2: Relate to geometry
Depth of bulb, \( h = 80 \, \text{cm} \).
Radius of circular area, \( r = h \tan i_c \).
Find \( i_c \) from \( \sin i_c \), then find \( \tan i_c \).
Step 3: Calculate area
Area = \( \pi r^2 = \pi (h \tan i_c)^2 \).
Substitute values to get area in cm² or m².
Question 9.6
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer & Explanation:
Step 1: Use prism formula for air
For minimum deviation, \( n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \).
Given \( A = 60^\circ \), \( \delta_m = 40^\circ \).
Calculate \( n_{\text{glass}} \).
Step 2: Prism in water
Relative refractive index of glass w.r.t water: \( n_{\text{gw}} = \frac{n_g}{n_w} = \frac{n_{\text{glass}}}{1.33} \).
Step 3: Find new minimum deviation (\( \delta_m' \))
\( n_{\text{gw}} = \frac{\sin\left(\frac{60^\circ + \delta_m'}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \).
Solve for \( \delta_m' \) using known \( n_{\text{gw}} \).
For minimum deviation, \( n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \).
Given \( A = 60^\circ \), \( \delta_m = 40^\circ \).
Calculate \( n_{\text{glass}} \).
Step 2: Prism in water
Relative refractive index of glass w.r.t water: \( n_{\text{gw}} = \frac{n_g}{n_w} = \frac{n_{\text{glass}}}{1.33} \).
Step 3: Find new minimum deviation (\( \delta_m' \))
\( n_{\text{gw}} = \frac{\sin\left(\frac{60^\circ + \delta_m'}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \).
Solve for \( \delta_m' \) using known \( n_{\text{gw}} \).
Question 9.8
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Answer & Explanation:
Key Concept: For a converging incident beam, the object is virtual. The point towards which the beam would converge without the lens acts as a virtual object for the lens.
Step 1: Sign convention
Distance of virtual object from lens, \( u = +12 \, \text{cm} \) (positive because object is on the side of incoming light).
Step 2: (a) For convex lens, \( f = +20 \, \text{cm} \)
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{12} = \frac{1}{20} \).
Solve for \( v \). Positive \( v \) means real image on the other side.
Step 3: (b) For concave lens, \( f = -16 \, \text{cm} \)
\( \frac{1}{v} - \frac{1}{12} = \frac{1}{-16} \).
Solve for \( v \). Negative \( v \) means virtual image on the same side as the incoming light.
Step 1: Sign convention
Distance of virtual object from lens, \( u = +12 \, \text{cm} \) (positive because object is on the side of incoming light).
Step 2: (a) For convex lens, \( f = +20 \, \text{cm} \)
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{12} = \frac{1}{20} \).
Solve for \( v \). Positive \( v \) means real image on the other side.
Step 3: (b) For concave lens, \( f = -16 \, \text{cm} \)
\( \frac{1}{v} - \frac{1}{12} = \frac{1}{-16} \).
Solve for \( v \). Negative \( v \) means virtual image on the same side as the incoming light.
Question 9.9
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Answer & Explanation:
Step 1: Find image distance (v)
For concave lens: \( f = -21 \, \text{cm} \), \( u = -14 \, \text{cm} \).
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{-14} = \frac{1}{-21} \).
Solve for \( v \). It will be negative, indicating virtual image on same side as object.
Step 2: Find magnification and size
\( m = \frac{v}{u} \). Calculate m. It will be positive and less than 1, indicating erect, diminished image.
Image size = \( m \times \text{object size} = m \times 3.0 \, \text{cm} \).
Step 3: Object moved further away
As \( |u| \) increases, \( |v| \) approaches \( |f| \) from the lens side. Image remains virtual, erect, and diminished, moving closer to the focus.
For concave lens: \( f = -21 \, \text{cm} \), \( u = -14 \, \text{cm} \).
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{-14} = \frac{1}{-21} \).
Solve for \( v \). It will be negative, indicating virtual image on same side as object.
Step 2: Find magnification and size
\( m = \frac{v}{u} \). Calculate m. It will be positive and less than 1, indicating erect, diminished image.
Image size = \( m \times \text{object size} = m \times 3.0 \, \text{cm} \).
Step 3: Object moved further away
As \( |u| \) increases, \( |v| \) approaches \( |f| \) from the lens side. Image remains virtual, erect, and diminished, moving closer to the focus.
Question 9.10
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer & Explanation:
Step 1: Formula for combination
For lenses in contact: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).
Convex lens: \( f_1 = +30 \, \text{cm} \).
Concave lens: \( f_2 = -20 \, \text{cm} \).
Step 2: Calculate equivalent focal length (F)
\( \frac{1}{F} = \frac{1}{30} + \frac{1}{(-20)} = \frac{1}{30} - \frac{1}{20} \).
\( \frac{1}{F} = \frac{2 - 3}{60} = -\frac{1}{60} \).
\( F = -60 \, \text{cm} \).
Step 3: Nature of combination
Negative focal length indicates the combination acts as a diverging (concave) lens of focal length 60 cm.
For lenses in contact: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).
Convex lens: \( f_1 = +30 \, \text{cm} \).
Concave lens: \( f_2 = -20 \, \text{cm} \).
Step 2: Calculate equivalent focal length (F)
\( \frac{1}{F} = \frac{1}{30} + \frac{1}{(-20)} = \frac{1}{30} - \frac{1}{20} \).
\( \frac{1}{F} = \frac{2 - 3}{60} = -\frac{1}{60} \).
\( F = -60 \, \text{cm} \).
Step 3: Nature of combination
Negative focal length indicates the combination acts as a diverging (concave) lens of focal length 60 cm.
Question 9.11
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) infinity? What is the magnifying power of the microscope in each case?
Answer & Explanation:
Given: \( f_o = 2.0 \, \text{cm} \), \( f_e = 6.25 \, \text{cm} \), Length L ≈ distance between lenses = 15 cm (approx).
(a) Final image at D = 25 cm (for eyepiece)
Step 1: Find object distance for eyepiece (\( u_e \))
For eyepiece: \( v_e = -25 \, \text{cm} \) (virtual image), \( f_e = +6.25 \, \text{cm} \).
Lens formula for eyepiece: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
Solve for \( u_e \). It will be negative.
Step 2: Find image distance for objective (\( v_o \))
Distance between lenses \( \approx v_o + |u_e| \).
So, \( v_o = L - |u_e| \).
Step 3: Find object distance for objective (\( u_o \))
For objective: \( f_o = +2.0 \, \text{cm} \), \( v_o \) from step 2.
Use lens formula: \( \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \). Solve for \( u_o \) (will be negative).
Step 4: Magnifying power (m)
\( m = m_o \times m_e = \left( \frac{v_o}{|u_o|} \right) \times \left( 1 + \frac{D}{f_e} \right) \).
Calculate m.
(b) Final image at infinity
For eyepiece: Object at its focus, so \( |u_e| = f_e = 6.25 \, \text{cm} \).
Then \( v_o = L - f_e = 15 - 6.25 = 8.75 \, \text{cm} \).
Find \( u_o \) using lens formula for objective as before.
Magnifying power: \( m = \left( \frac{v_o}{|u_o|} \right) \times \left( \frac{D}{f_e} \right) \).
(a) Final image at D = 25 cm (for eyepiece)
Step 1: Find object distance for eyepiece (\( u_e \))
For eyepiece: \( v_e = -25 \, \text{cm} \) (virtual image), \( f_e = +6.25 \, \text{cm} \).
Lens formula for eyepiece: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
Solve for \( u_e \). It will be negative.
Step 2: Find image distance for objective (\( v_o \))
Distance between lenses \( \approx v_o + |u_e| \).
So, \( v_o = L - |u_e| \).
Step 3: Find object distance for objective (\( u_o \))
For objective: \( f_o = +2.0 \, \text{cm} \), \( v_o \) from step 2.
Use lens formula: \( \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \). Solve for \( u_o \) (will be negative).
Step 4: Magnifying power (m)
\( m = m_o \times m_e = \left( \frac{v_o}{|u_o|} \right) \times \left( 1 + \frac{D}{f_e} \right) \).
Calculate m.
(b) Final image at infinity
For eyepiece: Object at its focus, so \( |u_e| = f_e = 6.25 \, \text{cm} \).
Then \( v_o = L - f_e = 15 - 6.25 = 8.75 \, \text{cm} \).
Find \( u_o \) using lens formula for objective as before.
Magnifying power: \( m = \left( \frac{v_o}{|u_o|} \right) \times \left( \frac{D}{f_e} \right) \).
Question 9.12
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer & Explanation:
Given: \( f_o = 8.0 \, \text{mm} = 0.8 \, \text{cm} \), \( f_e = 2.5 \, \text{cm} \), \( u_o = -9.0 \, \text{mm} = -0.9 \, \text{cm} \), \( D = 25 \, \text{cm} \).
Step 1: Find image distance for objective (\( v_o \))
Use lens formula for objective: \( \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \).
\( \frac{1}{v_o} - \frac{1}{-0.9} = \frac{1}{0.8} \).
Solve for \( v_o \). It will be positive.
Step 2: Assume final image at D (25 cm) for eyepiece
For eyepiece: \( v_e = -25 \, \text{cm} \), \( f_e = +2.5 \, \text{cm} \).
Find \( u_e \) using lens formula: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
Step 3: Find separation between lenses (L)
Separation \( L = v_o + |u_e| \). (since \( u_e \) is negative for eyepiece object).
Step 4: Calculate magnifying power (m)
\( m = m_o \times m_e = \left( \frac{v_o}{|u_o|} \right) \times \left( 1 + \frac{D}{f_e} \right) \).
Substitute values to get m.
Step 1: Find image distance for objective (\( v_o \))
Use lens formula for objective: \( \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \).
\( \frac{1}{v_o} - \frac{1}{-0.9} = \frac{1}{0.8} \).
Solve for \( v_o \). It will be positive.
Step 2: Assume final image at D (25 cm) for eyepiece
For eyepiece: \( v_e = -25 \, \text{cm} \), \( f_e = +2.5 \, \text{cm} \).
Find \( u_e \) using lens formula: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
Step 3: Find separation between lenses (L)
Separation \( L = v_o + |u_e| \). (since \( u_e \) is negative for eyepiece object).
Step 4: Calculate magnifying power (m)
\( m = m_o \times m_e = \left( \frac{v_o}{|u_o|} \right) \times \left( 1 + \frac{D}{f_e} \right) \).
Substitute values to get m.
Question 9.13
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece when the final image is formed at infinity?
Answer & Explanation:
Given: \( f_o = 144 \, \text{cm} \), \( f_e = 6.0 \, \text{cm} \).
Step 1: Magnifying power for normal adjustment (image at infinity)
\( m = \frac{f_o}{f_e} = \frac{144}{6.0} = 24 \).
Step 2: Separation between lenses for normal adjustment
When final image at infinity, object for eyepiece is at its focal point.
Image formed by objective lies at focus of eyepiece.
Separation \( L = f_o + f_e = 144 + 6.0 = 150 \, \text{cm} \).
Step 1: Magnifying power for normal adjustment (image at infinity)
\( m = \frac{f_o}{f_e} = \frac{144}{6.0} = 24 \).
Step 2: Separation between lenses for normal adjustment
When final image at infinity, object for eyepiece is at its focal point.
Image formed by objective lies at focus of eyepiece.
Separation \( L = f_o + f_e = 144 + 6.0 = 150 \, \text{cm} \).
Question 9.14
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is \( 3.48 \times 10^6 \, \text{m} \), and the radius of lunar orbit is \( 3.8 \times 10^8 \, \text{m} \).
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is \( 3.48 \times 10^6 \, \text{m} \), and the radius of lunar orbit is \( 3.8 \times 10^8 \, \text{m} \).
Answer & Explanation:
(a) Angular magnification
\( f_o = 15 \, \text{m} = 1500 \, \text{cm} \), \( f_e = 1.0 \, \text{cm} \).
\( m = \frac{f_o}{f_e} = \frac{1500}{1.0} = 1500 \).
(b) Diameter of moon's image by objective
Objective forms a real image at its focal plane.
Angle subtended by moon at objective ≈ \( \alpha = \frac{\text{Diameter of moon}}{\text{Distance to moon}} = \frac{3.48 \times 10^6}{3.8 \times 10^8} \, \text{radians} \).
This angle is very small.
Diameter of image = \( \alpha \times f_o \).
Calculate: Diameter = \( \left( \frac{3.48 \times 10^6}{3.8 \times 10^8} \right) \times 15 \, \text{m} \).
Convert to cm if needed.
\( f_o = 15 \, \text{m} = 1500 \, \text{cm} \), \( f_e = 1.0 \, \text{cm} \).
\( m = \frac{f_o}{f_e} = \frac{1500}{1.0} = 1500 \).
(b) Diameter of moon's image by objective
Objective forms a real image at its focal plane.
Angle subtended by moon at objective ≈ \( \alpha = \frac{\text{Diameter of moon}}{\text{Distance to moon}} = \frac{3.48 \times 10^6}{3.8 \times 10^8} \, \text{radians} \).
This angle is very small.
Diameter of image = \( \alpha \times f_o \).
Calculate: Diameter = \( \left( \frac{3.48 \times 10^6}{3.8 \times 10^8} \right) \times 15 \, \text{m} \).
Convert to cm if needed.
Question 9.15
Use the mirror equation to deduce that:
(a) an object placed between \( f \) and \( 2f \) of a concave mirror produces a real image beyond \( 2f \).
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
(a) an object placed between \( f \) and \( 2f \) of a concave mirror produces a real image beyond \( 2f \).
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Answer & Explanation:
Use mirror formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) with sign conventions: concave mirror \( f < 0 \), convex mirror \( f > 0 \), real object \( u < 0 \).
(a) For concave mirror: Let \( f = -F \), \( u = -u' \) where \( F < u' < 2F \).
Substitute in formula and solve for \( v \). Show \( v < -2F \), i.e., real image beyond \( 2f \).
(b) For convex mirror: \( f > 0 \), \( u < 0 \).
From formula, \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \). Since \( 1/u \) is negative, \( 1/v > 1/f > 0 \), so \( v > 0 \) always. Positive \( v \) means virtual image behind mirror.
(c) For convex mirror: Magnification \( m = -\frac{v}{u} \). With \( v > 0 \) and \( u < 0 \), \( m > 0 \) but magnitude \( |m| = v/|u| \). From formula, show \( v < |u| \) always, so \( |m| < 1 \) (diminished). Also show \( v < f \), so image between pole and focus.
(d) For concave mirror: \( f = -F \), \( u = -u' \) with \( 0 < u' < F \).
Show \( v > 0 \) (virtual), and \( |v/u| > 1 \) (enlarged).
(a) For concave mirror: Let \( f = -F \), \( u = -u' \) where \( F < u' < 2F \).
Substitute in formula and solve for \( v \). Show \( v < -2F \), i.e., real image beyond \( 2f \).
(b) For convex mirror: \( f > 0 \), \( u < 0 \).
From formula, \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \). Since \( 1/u \) is negative, \( 1/v > 1/f > 0 \), so \( v > 0 \) always. Positive \( v \) means virtual image behind mirror.
(c) For convex mirror: Magnification \( m = -\frac{v}{u} \). With \( v > 0 \) and \( u < 0 \), \( m > 0 \) but magnitude \( |m| = v/|u| \). From formula, show \( v < |u| \) always, so \( |m| < 1 \) (diminished). Also show \( v < f \), so image between pole and focus.
(d) For concave mirror: \( f = -F \), \( u = -u' \) with \( 0 < u' < F \).
Show \( v > 0 \) (virtual), and \( |v/u| > 1 \) (enlarged).
Question 9.16
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab with refractive index 1.5? Does the answer depend on the location of the slab?
Answer & Explanation:
Step 1: Normal shift formula
When an object is viewed through a slab, apparent shift = Real thickness \( \times \left( 1 - \frac{1}{n} \right) \).
Here, thickness \( t = 15 \, \text{cm} \), \( n = 1.5 \).
Step 2: Calculate shift
Shift \( S = t \left( 1 - \frac{1}{n} \right) = 15 \left( 1 - \frac{1}{1.5} \right) \).
\( S = 15 \left( 1 - \frac{2}{3} \right) = 15 \times \frac{1}{3} = 5 \, \text{cm} \).
Step 3: Does it depend on slab location?
The normal shift is independent of the distance of the observer or the location of the slab, as long as the viewing is near normal incidence. The pin appears raised by 5 cm.
When an object is viewed through a slab, apparent shift = Real thickness \( \times \left( 1 - \frac{1}{n} \right) \).
Here, thickness \( t = 15 \, \text{cm} \), \( n = 1.5 \).
Step 2: Calculate shift
Shift \( S = t \left( 1 - \frac{1}{n} \right) = 15 \left( 1 - \frac{1}{1.5} \right) \).
\( S = 15 \left( 1 - \frac{2}{3} \right) = 15 \times \frac{1}{3} = 5 \, \text{cm} \).
Step 3: Does it depend on slab location?
The normal shift is independent of the distance of the observer or the location of the slab, as long as the viewing is near normal incidence. The pin appears raised by 5 cm.
Question 9.17
(a) Figure shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total internal reflection inside the pipe takes place?
(b) What is the answer if there is no outer covering and the pipe is in air?
(b) What is the answer if there is no outer covering and the pipe is in air?
Answer & Explanation:
Concept: For light to propagate through fibre by TIR, angle of incidence at core-cladding interface must be greater than critical angle.
(a) With cladding (ncore = 1.68, nclad = 1.44)
Critical angle at interface: \( \sin i_c = \frac{n_{\text{clad}}}{n_{\text{core}}} = \frac{1.44}{1.68} \).
Calculate \( i_c \).
For TIR inside core, angle inside core with normal to wall \( r' > i_c \).
At entrance face: \( n_{\text{air}} \sin i = n_{\text{core}} \sin r \), and \( r' = 90^\circ - r \).
Condition \( r' \ge i_c \) leads to \( \sin i \le n_{\text{core}} \cos i_c \).
Find maximum \( i_{\text{max}} \). Range is from 0 to \( i_{\text{max}} \) with axis.
(b) No cladding (air, n = 1)
\( \sin i_c' = \frac{1}{1.68} \). Calculate new \( i_c' \).
Repeat steps to find new \( i_{\text{max}}' \). Range is smaller.
(a) With cladding (ncore = 1.68, nclad = 1.44)
Critical angle at interface: \( \sin i_c = \frac{n_{\text{clad}}}{n_{\text{core}}} = \frac{1.44}{1.68} \).
Calculate \( i_c \).
For TIR inside core, angle inside core with normal to wall \( r' > i_c \).
At entrance face: \( n_{\text{air}} \sin i = n_{\text{core}} \sin r \), and \( r' = 90^\circ - r \).
Condition \( r' \ge i_c \) leads to \( \sin i \le n_{\text{core}} \cos i_c \).
Find maximum \( i_{\text{max}} \). Range is from 0 to \( i_{\text{max}} \) with axis.
(b) No cladding (air, n = 1)
\( \sin i_c' = \frac{1}{1.68} \). Calculate new \( i_c' \).
Repeat steps to find new \( i_{\text{max}}' \). Range is smaller.
Question 9.18
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we 'see' a virtual image, we are obviously bringing it on to the 'screen' (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we 'see' a virtual image, we are obviously bringing it on to the 'screen' (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Answer & Explanation:
(a) Yes. If a virtual object (converging incident rays) is placed in front of a convex mirror, it can produce a real image. Similarly, a plane mirror can produce a real image if virtual object is placed in front of it (though practically rare).
(b) No contradiction. "Cannot be caught on a screen" means the light rays do not actually meet; they appear to diverge from the image location. Our eye lens converges these diverging rays to form a real image on the retina.
(c) The fisherman looks taller. The diver observes the fisherman from water (denser) to air (rarer). Light from fisherman bends away from normal as it enters water, making the fisherman appear taller.
(d) Yes, apparent depth decreases when viewed obliquely. The formula \( d_{\text{app}} = \frac{d_{\text{real}}}{n} \) is for normal viewing. For oblique viewing, the apparent depth is less than this value.
(e) Yes. High refractive index (\( n \approx 2.42 \)) gives a very small critical angle (\( \sim 24.4^\circ \)). A diamond cutter exploits this by cutting facets such that most light entering undergoes TIR and emerges from the top, causing brilliance.
(b) No contradiction. "Cannot be caught on a screen" means the light rays do not actually meet; they appear to diverge from the image location. Our eye lens converges these diverging rays to form a real image on the retina.
(c) The fisherman looks taller. The diver observes the fisherman from water (denser) to air (rarer). Light from fisherman bends away from normal as it enters water, making the fisherman appear taller.
(d) Yes, apparent depth decreases when viewed obliquely. The formula \( d_{\text{app}} = \frac{d_{\text{real}}}{n} \) is for normal viewing. For oblique viewing, the apparent depth is less than this value.
(e) Yes. High refractive index (\( n \approx 2.42 \)) gives a very small critical angle (\( \sim 24.4^\circ \)). A diamond cutter exploits this by cutting facets such that most light entering undergoes TIR and emerges from the top, causing brilliance.
Question 9.19
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Answer & Explanation:
Concept: For a real image to be formed on a screen at a fixed distance from object, the lens equation must have a real solution for given object and image distances.
Let distance between object and image screen be \( D = 3 \, \text{m} \). Let object distance be \( -u \) (negative), image distance be \( v \) (positive), with \( u + v = D \).
Lens formula: \( \frac{1}{v} - \frac{1}{-u} = \frac{1}{f} \) → \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).
But \( v = D - u \), so \( \frac{1}{D-u} + \frac{1}{u} = \frac{1}{f} \).
Solve for \( f \) in terms of u: \( f = \frac{u(D-u)}{D} \).
For a real image, both u and v must be positive and less than D, so \( 0 < u < D \).
The expression \( f = \frac{u(D-u)}{D} \) is maximum when \( u = D/2 \).
Then \( f_{\text{max}} = \frac{(D/2)(D/2)}{D} = \frac{D}{4} \).
So, maximum possible focal length \( = \frac{3}{4} = 0.75 \, \text{m} = 75 \, \text{cm} \).
Let distance between object and image screen be \( D = 3 \, \text{m} \). Let object distance be \( -u \) (negative), image distance be \( v \) (positive), with \( u + v = D \).
Lens formula: \( \frac{1}{v} - \frac{1}{-u} = \frac{1}{f} \) → \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).
But \( v = D - u \), so \( \frac{1}{D-u} + \frac{1}{u} = \frac{1}{f} \).
Solve for \( f \) in terms of u: \( f = \frac{u(D-u)}{D} \).
For a real image, both u and v must be positive and less than D, so \( 0 < u < D \).
The expression \( f = \frac{u(D-u)}{D} \) is maximum when \( u = D/2 \).
Then \( f_{\text{max}} = \frac{(D/2)(D/2)}{D} = \frac{D}{4} \).
So, maximum possible focal length \( = \frac{3}{4} = 0.75 \, \text{m} = 75 \, \text{cm} \).
Question 9.20
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer & Explanation:
Given: Distance between object and screen, \( D = 90 \, \text{cm} \).
The lens can be placed at two positions where sharp image is formed (conjugate positions). Separation between these positions, \( d = 20 \, \text{cm} \).
Step 1: Relation from displacement method
If focal length is \( f \) and \( D > 4f \), there are two positions.
Distance between object and screen, \( D = u + v \).
Separation between two lens positions, \( d = |u_1 - u_2| \).
It can be shown that focal length \( f = \frac{D^2 - d^2}{4D} \).
Step 2: Substitute values
\( D = 90 \, \text{cm} \), \( d = 20 \, \text{cm} \).
\( f = \frac{(90)^2 - (20)^2}{4 \times 90} = \frac{8100 - 400}{360} = \frac{7700}{360} \).
Simplify to get \( f \approx 21.39 \, \text{cm} \).
The lens can be placed at two positions where sharp image is formed (conjugate positions). Separation between these positions, \( d = 20 \, \text{cm} \).
Step 1: Relation from displacement method
If focal length is \( f \) and \( D > 4f \), there are two positions.
Distance between object and screen, \( D = u + v \).
Separation between two lens positions, \( d = |u_1 - u_2| \).
It can be shown that focal length \( f = \frac{D^2 - d^2}{4D} \).
Step 2: Substitute values
\( D = 90 \, \text{cm} \), \( d = 20 \, \text{cm} \).
\( f = \frac{(90)^2 - (20)^2}{4 \times 90} = \frac{8100 - 400}{360} = \frac{7700}{360} \).
Simplify to get \( f \approx 21.39 \, \text{cm} \).
Question 9.21
(a) Determine the 'effective focal length' of the combination of two lenses in contact, if they are separated by a small distance. Explain physically.
(b) A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. What is the focal length of the combination? Is it converging or diverging?
(c) Calculate the focal length of the combination if the lenses in (b) are separated by a distance of 10 cm.
(b) A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. What is the focal length of the combination? Is it converging or diverging?
(c) Calculate the focal length of the combination if the lenses in (b) are separated by a distance of 10 cm.
Answer & Explanation:
(a) General formula for two lenses separated by distance d
Power of combination: \( P = P_1 + P_2 - d P_1 P_2 \), where \( P = 1/f \).
So, \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \).
Physically, separation reduces the effective power because the second lens does not receive light converged/diver
Power of combination: \( P = P_1 + P_2 - d P_1 P_2 \), where \( P = 1/f \).
So, \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \).
Physically, separation reduces the effective power because the second lens does not receive light converged/diver
Question 9.22
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Answer & Explanation:
Step 1: Condition for TIR at second face
For TIR at second face: \( r_2 \ge i_c \), where \( \sin i_c = \frac{1}{n} = \frac{1}{1.524} \).
Calculate \( i_c \).
Step 2: Relation between angles in prism
\( A = r_1 + r_2 \), where \( A = 60^\circ \).
For just TIR, \( r_2 = i_c \). So, \( r_1 = A - i_c \).
Step 3: Apply Snell's law at first face
\( \sin i_1 = n \sin r_1 \).
Substitute \( r_1 = 60^\circ - i_c \).
Calculate \( i_1 \). This is the required angle of incidence.
For TIR at second face: \( r_2 \ge i_c \), where \( \sin i_c = \frac{1}{n} = \frac{1}{1.524} \).
Calculate \( i_c \).
Step 2: Relation between angles in prism
\( A = r_1 + r_2 \), where \( A = 60^\circ \).
For just TIR, \( r_2 = i_c \). So, \( r_1 = A - i_c \).
Step 3: Apply Snell's law at first face
\( \sin i_1 = n \sin r_1 \).
Substitute \( r_1 = 60^\circ - i_c \).
Calculate \( i_1 \). This is the required angle of incidence.
Question 9.23
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion, (b) disperse (and displace) a pencil of white light without much deviation.
Answer & Explanation:
Concepts: Dispersion without deviation, deviation without dispersion using combination of prisms of different materials.
(a) Deviate without much dispersion (Achromatic Prism)
Combine a crown glass prism with a flint glass prism of smaller angle, placed opposite. Flint glass has higher dispersive power.
Choose angles so that the mean deviations are equal and opposite, while dispersion (angular) cancels.
Condition: \( (n_y - 1)A + (n'_y - 1)A' = 0 \) for mean deviation, and \( (n_v - n_r)A + (n'_v - n'_r)A' = 0 \) for achromatism.
(b) Disperse without much deviation (Direct Vision Prism)
Combine crown and flint glass prisms with bases opposite. Arrange so that mean deviation is zero, but dispersion adds up.
Condition: \( (n_y - 1)A = -(n'_y - 1)A' \) (zero mean deviation).
Then net dispersion = \( (n_v - n_r)A + (n'_v - n'_r)A' \) is non-zero.
(a) Deviate without much dispersion (Achromatic Prism)
Combine a crown glass prism with a flint glass prism of smaller angle, placed opposite. Flint glass has higher dispersive power.
Choose angles so that the mean deviations are equal and opposite, while dispersion (angular) cancels.
Condition: \( (n_y - 1)A + (n'_y - 1)A' = 0 \) for mean deviation, and \( (n_v - n_r)A + (n'_v - n'_r)A' = 0 \) for achromatism.
(b) Disperse without much deviation (Direct Vision Prism)
Combine crown and flint glass prisms with bases opposite. Arrange so that mean deviation is zero, but dispersion adds up.
Condition: \( (n_y - 1)A = -(n'_y - 1)A' \) (zero mean deviation).
Then net dispersion = \( (n_v - n_r)A + (n'_v - n'_r)A' \) is non-zero.
Question 9.24
For a normal eye, the far point is at infinity and the near point is at 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Answer & Explanation:
Step 1: Total power for far point
For far point at infinity, image forms on retina. Eye is relaxed, lens has minimum power \( P_{\text{lens min}} = 20 \, D \).
Cornea power \( P_{\text{cornea}} = 40 \, D \).
Total power \( P_{\text{far}} = 40 + 20 = 60 \, D \).
Step 2: Total power for near point
For near point at \( u = -25 \, \text{cm} = -0.25 \, \text{m} \), image on retina (distance from lens to retina ≈ focal length for far point).
Let distance from eye-lens to retina = \( v \). Using \( P_{\text{far}} = 60 \, D \), \( v = 1/60 \, \text{m} \approx 0.01667 \, \text{m} = 1.667 \, \text{cm} \).
For near object: \( u = -0.25 \, \text{m} \), \( v = 0.01667 \, \text{m} \).
Power needed by eye alone: \( P_{\text{total}} = \frac{1}{v} - \frac{1}{u} = \frac{1}{0.01667} - \frac{1}{-0.25} \approx 60 + 4 = 64 \, D \).
This total includes cornea. So lens power needed, \( P_{\text{lens max}} = 64 - 40 = 24 \, D \).
Step 3: Range of accommodation
Range = \( P_{\text{lens max}} - P_{\text{lens min}} = 24 - 20 = 4 \, D \).
For far point at infinity, image forms on retina. Eye is relaxed, lens has minimum power \( P_{\text{lens min}} = 20 \, D \).
Cornea power \( P_{\text{cornea}} = 40 \, D \).
Total power \( P_{\text{far}} = 40 + 20 = 60 \, D \).
Step 2: Total power for near point
For near point at \( u = -25 \, \text{cm} = -0.25 \, \text{m} \), image on retina (distance from lens to retina ≈ focal length for far point).
Let distance from eye-lens to retina = \( v \). Using \( P_{\text{far}} = 60 \, D \), \( v = 1/60 \, \text{m} \approx 0.01667 \, \text{m} = 1.667 \, \text{cm} \).
For near object: \( u = -0.25 \, \text{m} \), \( v = 0.01667 \, \text{m} \).
Power needed by eye alone: \( P_{\text{total}} = \frac{1}{v} - \frac{1}{u} = \frac{1}{0.01667} - \frac{1}{-0.25} \approx 60 + 4 = 64 \, D \).
This total includes cornea. So lens power needed, \( P_{\text{lens max}} = 64 - 40 = 24 \, D \).
Step 3: Range of accommodation
Range = \( P_{\text{lens max}} - P_{\text{lens min}} = 24 - 20 = 4 \, D \).
Question 9.25
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability to accommodate? If not, what might cause these defects of vision?
Answer & Explanation:
No, myopia and hypermetropia do not necessarily imply loss of accommodation ability. They are primarily refractive defects due to mismatch between eyeball length and refractive power.
1. Myopia (Short-sightedness):
• Cause: Eyeball too long, or cornea/lens too powerful.
• Far point comes closer (not infinity).
• Accommodation may be normal for near objects.
2. Hypermetropia (Long-sightedness):
• Cause: Eyeball too short, or cornea/lens too weak.
• Near point recedes beyond 25 cm.
• Accommodation may be normal but has to work harder for near vision.
Loss of accommodation (presbyopia) is a separate defect due to stiffening of eye-lens with age. Myopia/hypermetropia can exist with or without presbyopia.
1. Myopia (Short-sightedness):
• Cause: Eyeball too long, or cornea/lens too powerful.
• Far point comes closer (not infinity).
• Accommodation may be normal for near objects.
2. Hypermetropia (Long-sightedness):
• Cause: Eyeball too short, or cornea/lens too weak.
• Near point recedes beyond 25 cm.
• Accommodation may be normal but has to work harder for near vision.
Loss of accommodation (presbyopia) is a separate defect due to stiffening of eye-lens with age. Myopia/hypermetropia can exist with or without presbyopia.
Question 9.26
A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age he also needs to use separate reading glasses of power +2.0 dioptres. Explain what may have happened.
Answer & Explanation:
Step 1: Understanding the initial condition
Power of spectacles for distant vision = –1.0 D (concave lens).
This corrects myopia. Without glasses, far point is at some finite distance.
Using lens formula: \( \frac{1}{f} = P = -1.0 \, \text{m}^{-1} \), so \( f = -1.0 \, \text{m} = -100 \, \text{cm} \).
For a myopic eye, far point is at \( -f = 100 \, \text{cm} \) in front.
Step 2: What happened in old age?
The person develops presbyopia – loss of accommodation due to stiffening of eye-lens.
Even though myopia persists (far point still at 100 cm), the eye cannot increase its power sufficiently to see near objects clearly.
Step 3: Need for reading glasses
For reading at normal distance (say 25 cm), the eye needs extra converging power.
Reading glasses of +2.0 D provide this additional power.
Note: The person might use bifocals: upper part –1.0 D for distance, lower part +1.0 D (since myopia already gives some help for near) or +2.0 D depending on exact requirement.
Power of spectacles for distant vision = –1.0 D (concave lens).
This corrects myopia. Without glasses, far point is at some finite distance.
Using lens formula: \( \frac{1}{f} = P = -1.0 \, \text{m}^{-1} \), so \( f = -1.0 \, \text{m} = -100 \, \text{cm} \).
For a myopic eye, far point is at \( -f = 100 \, \text{cm} \) in front.
Step 2: What happened in old age?
The person develops presbyopia – loss of accommodation due to stiffening of eye-lens.
Even though myopia persists (far point still at 100 cm), the eye cannot increase its power sufficiently to see near objects clearly.
Step 3: Need for reading glasses
For reading at normal distance (say 25 cm), the eye needs extra converging power.
Reading glasses of +2.0 D provide this additional power.
Note: The person might use bifocals: upper part –1.0 D for distance, lower part +1.0 D (since myopia already gives some help for near) or +2.0 D depending on exact requirement.
Question 9.27
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
Answer & Explanation:
Defect: This is astigmatism.
Cause: The cornea (or sometimes lens) has different curvatures (and hence different focal lengths) in different planes (e.g., vertical and horizontal).
Here, vertical lines are seen distinctly → focusing is proper for vertical plane. Horizontal lines are blurred → improper focusing in horizontal plane.
Correction: Using a cylindrical lens with appropriate axis.
A cylindrical lens has power in one plane only (say, in horizontal plane to correct horizontal focusing).
The axis of the cylindrical lens is placed perpendicular to the plane needing correction.
Cause: The cornea (or sometimes lens) has different curvatures (and hence different focal lengths) in different planes (e.g., vertical and horizontal).
Here, vertical lines are seen distinctly → focusing is proper for vertical plane. Horizontal lines are blurred → improper focusing in horizontal plane.
Correction: Using a cylindrical lens with appropriate axis.
A cylindrical lens has power in one plane only (say, in horizontal plane to correct horizontal focusing).
The axis of the cylindrical lens is placed perpendicular to the plane needing correction.
Question 9.28
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
(a) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Answer & Explanation:
Given: \( f = +5 \, \text{cm} \), \( D = 25 \, \text{cm} \) (normal near point).
(a) Closest and farthest distances for reading
For magnifying glass, object distance can vary to produce image at near point or infinity.
• Closest distance: When image is at near point (\( v = -25 \, \text{cm} \), virtual).
Use lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{-25} - \frac{1}{u_{\text{min}}} = \frac{1}{5} \). Solve for \( u_{\text{min}} \) (negative). Magnitude is the closest object distance.
• Farthest distance: When image at infinity (eye relaxed). Then object at focus: \( u_{\text{max}} = -f = -5 \, \text{cm} \).
So, farthest object distance = 5 cm from lens.
(b) Maximum and minimum angular magnification
Angular magnification \( M = \frac{D}{|u|} \) for image at near point, and \( M = \frac{D}{f} \) for image at infinity.
• Maximum M: When image at near point, \( M_{\text{max}} = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 6 \).
• Minimum M: When image at infinity, \( M_{\text{min}} = \frac{D}{f} = \frac{25}{5} = 5 \).
(a) Closest and farthest distances for reading
For magnifying glass, object distance can vary to produce image at near point or infinity.
• Closest distance: When image is at near point (\( v = -25 \, \text{cm} \), virtual).
Use lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{-25} - \frac{1}{u_{\text{min}}} = \frac{1}{5} \). Solve for \( u_{\text{min}} \) (negative). Magnitude is the closest object distance.
• Farthest distance: When image at infinity (eye relaxed). Then object at focus: \( u_{\text{max}} = -f = -5 \, \text{cm} \).
So, farthest object distance = 5 cm from lens.
(b) Maximum and minimum angular magnification
Angular magnification \( M = \frac{D}{|u|} \) for image at near point, and \( M = \frac{D}{f} \) for image at infinity.
• Maximum M: When image at near point, \( M_{\text{max}} = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 6 \).
• Minimum M: When image at infinity, \( M_{\text{min}} = \frac{D}{f} = \frac{25}{5} = 5 \).
Question 9.29
A card sheet divided into squares each of size 1 mm² is being viewed from a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Answer & Explanation:
Given: Object distance \( u = -9 \, \text{cm} \), focal length \( f = +10 \, \text{cm} \).
(a) Lateral magnification and area
• Find image distance \( v \) using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{-9} = \frac{1}{10} \). Solve for \( v \). It will be negative (virtual image).
• Lateral magnification \( m = \frac{v}{u} \) (or \( \frac{h_i}{h_o} \)). Calculate \( m \).
• Area of each square in image = \( m^2 \times \) original area (since linear magnification in both directions is m).
Original area = 1 mm² = 0.01 cm². So image area = \( m^2 \times 0.01 \, \text{cm}^2 \).
(b) Angular magnification (M)
For a simple microscope with image at infinity (relaxed eye), \( M = \frac{D}{f} = \frac{25}{10} = 2.5 \).
But here object is not at focus; it's at 9 cm. Angular magnification is \( \frac{\text{Angle with lens}}{\text{Angle without lens}} \approx \frac{D}{|u|} \) if image at near point, or appropriate formula.
Since lens is close to eye and image likely at near point for comfortable viewing, \( M \approx \frac{D}{|u|} = \frac{25}{9} \approx 2.78 \).
(c) Comparison
No, lateral magnification (a) is not equal to angular magnifying power (b).
Lateral magnification relates to size of image vs size of object.
Angular magnification relates to angle subtended at eye with vs without lens.
(a) Lateral magnification and area
• Find image distance \( v \) using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{v} - \frac{1}{-9} = \frac{1}{10} \). Solve for \( v \). It will be negative (virtual image).
• Lateral magnification \( m = \frac{v}{u} \) (or \( \frac{h_i}{h_o} \)). Calculate \( m \).
• Area of each square in image = \( m^2 \times \) original area (since linear magnification in both directions is m).
Original area = 1 mm² = 0.01 cm². So image area = \( m^2 \times 0.01 \, \text{cm}^2 \).
(b) Angular magnification (M)
For a simple microscope with image at infinity (relaxed eye), \( M = \frac{D}{f} = \frac{25}{10} = 2.5 \).
But here object is not at focus; it's at 9 cm. Angular magnification is \( \frac{\text{Angle with lens}}{\text{Angle without lens}} \approx \frac{D}{|u|} \) if image at near point, or appropriate formula.
Since lens is close to eye and image likely at near point for comfortable viewing, \( M \approx \frac{D}{|u|} = \frac{25}{9} \approx 2.78 \).
(c) Comparison
No, lateral magnification (a) is not equal to angular magnifying power (b).
Lateral magnification relates to size of image vs size of object.
Angular magnification relates to angle subtended at eye with vs without lens.
Question 9.30
(a) At what distance should the lens be held from the card sheet in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Answer & Explanation:
Given: \( f = 10 \, \text{cm} \), \( D = 25 \, \text{cm} \) (normal near point).
(a) Distance for maximum magnifying power
Maximum magnifying power occurs when image is at near point (\( v = -D = -25 \, \text{cm} \)).
Use lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{-25} - \frac{1}{u} = \frac{1}{10} \). Solve for \( u \) (object distance). It will be negative.
Distance from lens = \( |u| \).
(b) Magnification in this case
Lateral magnification \( m = \frac{v}{u} = \frac{-25}{u} \) (with u negative).
Or \( m = 1 + \frac{D}{f} = 1 + \frac{25}{10} = 3.5 \). (since \( m = \frac{D}{|u|} \) and also \( \frac{1}{|u|} = \frac{1}{f} + \frac{1}{D} \)).
(c) Is magnification equal to magnifying power?
Yes, in this specific case when image is at near point, the angular magnification (magnifying power) is \( 1 + \frac{D}{f} \), which equals the lateral magnification \( \frac{v}{u} \) for the conditions. They coincide.
(a) Distance for maximum magnifying power
Maximum magnifying power occurs when image is at near point (\( v = -D = -25 \, \text{cm} \)).
Use lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{-25} - \frac{1}{u} = \frac{1}{10} \). Solve for \( u \) (object distance). It will be negative.
Distance from lens = \( |u| \).
(b) Magnification in this case
Lateral magnification \( m = \frac{v}{u} = \frac{-25}{u} \) (with u negative).
Or \( m = 1 + \frac{D}{f} = 1 + \frac{25}{10} = 3.5 \). (since \( m = \frac{D}{|u|} \) and also \( \frac{1}{|u|} = \frac{1}{f} + \frac{1}{D} \)).
(c) Is magnification equal to magnifying power?
Yes, in this specific case when image is at near point, the angular magnification (magnifying power) is \( 1 + \frac{D}{f} \), which equals the lateral magnification \( \frac{v}{u} \) for the conditions. They coincide.
Question 9.31
What should be the distance between the object and the magnifying glass if the virtual image of each square in Exercise 9.29 is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Answer & Explanation:
Given: Original area of square = 1 mm².
Desired image area = 6.25 mm².
Focal length \( f = 10 \, \text{cm} \).
Step 1: Find lateral magnification (m)
Since area magnification = \( m^2 \), we have:
\( m^2 = \frac{6.25}{1} = 6.25 \) ⇒ \( m = 2.5 \) (positive for virtual erect image).
Step 2: Find object distance (u)
Magnification \( m = \frac{v}{u} \) (with sign: u negative, v negative for virtual image).
So, \( v = m u \).
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
Substitute \( v = 2.5u \): \( \frac{1}{2.5u} - \frac{1}{u} = \frac{1}{10} \).
Solve for u (will be negative). Find \( |u| \).
Step 3: Distinct vision?
Since image is virtual and magnified, if eyes are very close to magnifier, the image can be seen distinctly as long as it is within eye's near point (25 cm) or slightly beyond. Need to check v: \( v = 2.5u \). Calculate v; if \( |v| \) is around 25 cm or less, it's comfortable.
Desired image area = 6.25 mm².
Focal length \( f = 10 \, \text{cm} \).
Step 1: Find lateral magnification (m)
Since area magnification = \( m^2 \), we have:
\( m^2 = \frac{6.25}{1} = 6.25 \) ⇒ \( m = 2.5 \) (positive for virtual erect image).
Step 2: Find object distance (u)
Magnification \( m = \frac{v}{u} \) (with sign: u negative, v negative for virtual image).
So, \( v = m u \).
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
Substitute \( v = 2.5u \): \( \frac{1}{2.5u} - \frac{1}{u} = \frac{1}{10} \).
Solve for u (will be negative). Find \( |u| \).
Step 3: Distinct vision?
Since image is virtual and magnified, if eyes are very close to magnifier, the image can be seen distinctly as long as it is within eye's near point (25 cm) or slightly beyond. Need to check v: \( v = 2.5u \). Calculate v; if \( |v| \) is around 25 cm or less, it's comfortable.
Question 9.32
Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of very small focal length for greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of very small focal length for greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer & Explanation:
(a) The statement is true only when object is placed at focal point, producing image at infinity. Then angle subtended by object at eye placed at lens is same as angle subtended by image. But angular magnification is defined as ratio of angle subtended by image (with instrument) to angle subtended by object at near point (without instrument, placed at 25 cm). The magnifier allows object to be brought closer than 25 cm, increasing the angle.
(b) Yes, angular magnification changes slightly if eye is moved back. The angular size of image decreases as eye recedes from lens. Maximum angular magnification is achieved when eye is close to lens and image at near point.
(c) Practical limitations: Lenses with very small focal length have small radius of curvature, leading to severe aberrations (spherical, chromatic). Also, such lenses would have very small diameter, giving a tiny field of view.
(d) Short focal length of objective gives larger magnification (\( m_o = v_o / u_o \approx L/f_o \) for large L). Short focal length of eyepiece gives larger angular magnification (\( m_e \approx D/f_e \)). Together they give high overall magnification.
(e) Positioning eye slightly away allows the entire field of view (the circle of light emerging from eyepiece, called Ramsden circle or eye ring) to enter the eye. This maximizes brightness and field.
The short distance is roughly equal to the distance of the eye-ring from the eyepiece (typically a few mm to 1 cm).
(b) Yes, angular magnification changes slightly if eye is moved back. The angular size of image decreases as eye recedes from lens. Maximum angular magnification is achieved when eye is close to lens and image at near point.
(c) Practical limitations: Lenses with very small focal length have small radius of curvature, leading to severe aberrations (spherical, chromatic). Also, such lenses would have very small diameter, giving a tiny field of view.
(d) Short focal length of objective gives larger magnification (\( m_o = v_o / u_o \approx L/f_o \) for large L). Short focal length of eyepiece gives larger angular magnification (\( m_e \approx D/f_e \)). Together they give high overall magnification.
(e) Positioning eye slightly away allows the entire field of view (the circle of light emerging from eyepiece, called Ramsden circle or eye ring) to enter the eye. This maximizes brightness and field.
The short distance is roughly equal to the distance of the eye-ring from the eyepiece (typically a few mm to 1 cm).
Question 9.33
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Answer & Explanation:
Given: Desired magnifying power \( M = 30 \), \( f_o = 1.25 \, \text{cm} \), \( f_e = 5 \, \text{cm} \). Assume normal eye, image at infinity for comfortable viewing.
Step 1: Magnifying power formula (image at infinity)
\( M = m_o \times m_e = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right) \), where \( L \) = tube length (distance between focal points of objective and eyepiece), \( D = 25 \, \text{cm} \).
So, \( 30 = \left( \frac{L}{1.25} \right) \times \left( \frac{25}{5} \right) = \frac{L}{1.25} \times 5 \).
Solve for \( L \): \( L = \frac{30 \times 1.25}{5} = 7.5 \, \text{cm} \).
Step 2: Set up distances
• Place object just outside focus of objective: \( u_o \approx -f_o = -1.25 \, \text{cm} \) (slightly more).
• Image from objective forms at distance \( v_o \) from objective such that \( v_o \approx L + f_o = 7.5 + 1.25 = 8.75 \, \text{cm} \) (since \( L \approx v_o - f_o \)).
• This image acts as object for eyepiece, placed at its focus (\( u_e = -f_e = -5 \, \text{cm} \)) for image at infinity.
• Separation between lenses = \( v_o + |u_e| = 8.75 + 5 = 13.75 \, \text{cm} \).
Step 3: Adjustments
Fine-tune \( u_o \) slightly to get sharp image at infinity through eyepiece.
Step 1: Magnifying power formula (image at infinity)
\( M = m_o \times m_e = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right) \), where \( L \) = tube length (distance between focal points of objective and eyepiece), \( D = 25 \, \text{cm} \).
So, \( 30 = \left( \frac{L}{1.25} \right) \times \left( \frac{25}{5} \right) = \frac{L}{1.25} \times 5 \).
Solve for \( L \): \( L = \frac{30 \times 1.25}{5} = 7.5 \, \text{cm} \).
Step 2: Set up distances
• Place object just outside focus of objective: \( u_o \approx -f_o = -1.25 \, \text{cm} \) (slightly more).
• Image from objective forms at distance \( v_o \) from objective such that \( v_o \approx L + f_o = 7.5 + 1.25 = 8.75 \, \text{cm} \) (since \( L \approx v_o - f_o \)).
• This image acts as object for eyepiece, placed at its focus (\( u_e = -f_e = -5 \, \text{cm} \)) for image at infinity.
• Separation between lenses = \( v_o + |u_e| = 8.75 + 5 = 13.75 \, \text{cm} \).
Step 3: Adjustments
Fine-tune \( u_o \) slightly to get sharp image at infinity through eyepiece.
Question 9.34
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment (i.e., when the final image is at infinity)?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
Answer & Explanation:
Part 1: Magnifying power
\( f_o = 140 \, \text{cm} \), \( f_e = 5.0 \, \text{cm} \).
For normal adjustment (image at infinity):
\( M = \frac{f_o}{f_e} = \frac{140}{5.0} = 28 \).
Part 2: Height of image by objective
Tower height \( h_o = 100 \, \text{m} \), distance from objective \( u_o \approx -3 \, \text{km} = -3000 \, \text{m} \) (negative).
Objective forms real image at its focal plane (since object distant).
Angle subtended by tower: \( \theta \approx \frac{h_o}{\text{distance}} = \frac{100}{3000} = \frac{1}{30} \, \text{radians} \).
Height of image \( h_i = \theta \times f_o \) (since \( \theta \approx \frac{h_i}{f_o} \)).
\( f_o = 140 \, \text{cm} = 1.4 \, \text{m} \).
So, \( h_i = \frac{1}{30} \times 1.4 \approx 0.04667 \, \text{m} = 4.67 \, \text{cm} \).
\( f_o = 140 \, \text{cm} \), \( f_e = 5.0 \, \text{cm} \).
For normal adjustment (image at infinity):
\( M = \frac{f_o}{f_e} = \frac{140}{5.0} = 28 \).
Part 2: Height of image by objective
Tower height \( h_o = 100 \, \text{m} \), distance from objective \( u_o \approx -3 \, \text{km} = -3000 \, \text{m} \) (negative).
Objective forms real image at its focal plane (since object distant).
Angle subtended by tower: \( \theta \approx \frac{h_o}{\text{distance}} = \frac{100}{3000} = \frac{1}{30} \, \text{radians} \).
Height of image \( h_i = \theta \times f_o \) (since \( \theta \approx \frac{h_i}{f_o} \)).
\( f_o = 140 \, \text{cm} = 1.4 \, \text{m} \).
So, \( h_i = \frac{1}{30} \times 1.4 \approx 0.04667 \, \text{m} = 4.67 \, \text{cm} \).
Question 9.35
(a) For the telescope described in Exercise 9.34, what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the final image of the tower if the final image is formed at 25 cm?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the final image of the tower if the final image is formed at 25 cm?
Answer & Explanation:
Given: \( f_o = 140 \, \text{cm} \), \( f_e = 5.0 \, \text{cm} \).
(a) Separation in normal adjustment
When final image at infinity, object for eyepiece is at its focus.
Image from objective lies at focus of eyepiece.
Separation = \( f_o + f_e = 140 + 5.0 = 145 \, \text{cm} \).
(b) Height of final image (image at near point, D = 25 cm)
First, find image height by objective (as in Q9.34): \( h_i' = 4.67 \, \text{cm} \) (approx). This is object for eyepiece.
For eyepiece: \( f_e = 5.0 \, \text{cm} \), final image at \( v_e = -25 \, \text{cm} \) (virtual).
Find object distance for eyepiece \( u_e \) using lens formula: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
\( \frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \). Solve for \( u_e \) (negative).
Magnification of eyepiece \( m_e = \frac{v_e}{u_e} = \frac{-25}{u_e} \).
Final image height = \( h_i' \times |m_e| \).
(a) Separation in normal adjustment
When final image at infinity, object for eyepiece is at its focus.
Image from objective lies at focus of eyepiece.
Separation = \( f_o + f_e = 140 + 5.0 = 145 \, \text{cm} \).
(b) Height of final image (image at near point, D = 25 cm)
First, find image height by objective (as in Q9.34): \( h_i' = 4.67 \, \text{cm} \) (approx). This is object for eyepiece.
For eyepiece: \( f_e = 5.0 \, \text{cm} \), final image at \( v_e = -25 \, \text{cm} \) (virtual).
Find object distance for eyepiece \( u_e \) using lens formula: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \).
\( \frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \). Solve for \( u_e \) (negative).
Magnification of eyepiece \( m_e = \frac{v_e}{u_e} = \frac{-25}{u_e} \).
Final image height = \( h_i' \times |m_e| \).
Question 9.36
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror has a radius of curvature 140 mm, where will the final image of an object at infinity be?
Answer & Explanation:
Given: Cassegrain telescope: Large concave primary mirror (M1), small convex secondary mirror (M2).
Separation between mirrors = 20 mm.
Radius of curvature of M1, \( R_1 = 220 \, \text{mm} \) ⇒ focal length \( f_1 = R_1/2 = 110 \, \text{mm} \) (positive for concave mirror?). Wait: For mirror, \( f = R/2 \). For concave mirror (used as primary), \( f_1 = -110 \, \text{mm} \) (by sign convention, if direction of incident light is positive).
Radius of curvature of M2, \( R_2 = 140 \, \text{mm} \) (convex mirror) ⇒ \( f_2 = R_2/2 = 70 \, \text{mm} \), and since convex, \( f_2 = +70 \, \text{mm} \).
Step 1: Effect of primary mirror alone
Object at infinity → image at its focus, which is 110 mm in front of M1.
Step 2: This image acts as virtual object for secondary mirror
Distance of this image from M2: Since mirrors are 20 mm apart, the image from M1 is located (110 - 20) = 90 mm behind M2? Need careful sign convention.
Let's set coordinate: Light from infinity comes from left, hits M1 first. Take left to right as positive direction.
For M1 (concave): \( u_1 = \infty \), \( f_1 = -110 \, \text{mm} \). Mirror formula: \( \frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f_1} \).
\( \frac{1}{v_1} + 0 = \frac{1}{-110} \) ⇒ \( v_1 = -110 \, \text{mm} \). Negative means image is on same side as incoming light (left side of M1). So image I1 is 110 mm in front of M1.
Step 3: For M2 (convex)
M2 is 20 mm to the right of M1. So distance of I1 from M2 = \( -110 - 20 = -130 \, \text{mm} \) (since I1 is left of M1, and M2 is right of M1). So object distance for M2, \( u_2 = -130 \, \text{mm} \).
For convex mirror M2: \( f_2 = +70 \, \text{mm} \).
Mirror formula: \( \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2} \).
\( \frac{1}{v_2} + \frac{1}{-130} = \frac{1}{70} \).
Solve for \( v_2 \). It will be positive, meaning image is behind M2 (virtual for M2, but real for overall?).
This \( v_2 \) gives location of final image relative to M2.
Separation between mirrors = 20 mm.
Radius of curvature of M1, \( R_1 = 220 \, \text{mm} \) ⇒ focal length \( f_1 = R_1/2 = 110 \, \text{mm} \) (positive for concave mirror?). Wait: For mirror, \( f = R/2 \). For concave mirror (used as primary), \( f_1 = -110 \, \text{mm} \) (by sign convention, if direction of incident light is positive).
Radius of curvature of M2, \( R_2 = 140 \, \text{mm} \) (convex mirror) ⇒ \( f_2 = R_2/2 = 70 \, \text{mm} \), and since convex, \( f_2 = +70 \, \text{mm} \).
Step 1: Effect of primary mirror alone
Object at infinity → image at its focus, which is 110 mm in front of M1.
Step 2: This image acts as virtual object for secondary mirror
Distance of this image from M2: Since mirrors are 20 mm apart, the image from M1 is located (110 - 20) = 90 mm behind M2? Need careful sign convention.
Let's set coordinate: Light from infinity comes from left, hits M1 first. Take left to right as positive direction.
For M1 (concave): \( u_1 = \infty \), \( f_1 = -110 \, \text{mm} \). Mirror formula: \( \frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f_1} \).
\( \frac{1}{v_1} + 0 = \frac{1}{-110} \) ⇒ \( v_1 = -110 \, \text{mm} \). Negative means image is on same side as incoming light (left side of M1). So image I1 is 110 mm in front of M1.
Step 3: For M2 (convex)
M2 is 20 mm to the right of M1. So distance of I1 from M2 = \( -110 - 20 = -130 \, \text{mm} \) (since I1 is left of M1, and M2 is right of M1). So object distance for M2, \( u_2 = -130 \, \text{mm} \).
For convex mirror M2: \( f_2 = +70 \, \text{mm} \).
Mirror formula: \( \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2} \).
\( \frac{1}{v_2} + \frac{1}{-130} = \frac{1}{70} \).
Solve for \( v_2 \). It will be positive, meaning image is behind M2 (virtual for M2, but real for overall?).
This \( v_2 \) gives location of final image relative to M2.
Question 9.37
Light incident normally on a plane mirror attached to a galvanometer coil retraces the path after reflection. If a tiny deflection of the coil causes the reflected spot to move by 2 mm on a screen placed 1 m away, what is the deflection of the coil?
Answer & Explanation:
Concept: If a mirror rotates by an angle \( \theta \), the reflected ray rotates by \( 2\theta \).
Given: Spot displacement on screen \( y = 2 \, \text{mm} = 0.002 \, \text{m} \).
Distance from mirror to screen \( D = 1 \, \text{m} \).
Step 1: Find angular deflection of reflected ray
Since spot moves 2 mm on screen 1 m away, angle of rotation of reflected ray is
\( \alpha \approx \frac{y}{D} = \frac{0.002}{1} = 0.002 \, \text{radians} \).
Step 2: Relate to mirror rotation
If mirror rotates by \( \theta \), reflected ray rotates by \( 2\theta \).
So, \( 2\theta = \alpha \) ⇒ \( \theta = \frac{\alpha}{2} = \frac{0.002}{2} = 0.001 \, \text{radians} \).
Answer: Deflection of the coil = \( 0.001 \, \text{rad} \) or \( \frac{180}{\pi} \times 0.001 \approx 0.0573^\circ \).
Given: Spot displacement on screen \( y = 2 \, \text{mm} = 0.002 \, \text{m} \).
Distance from mirror to screen \( D = 1 \, \text{m} \).
Step 1: Find angular deflection of reflected ray
Since spot moves 2 mm on screen 1 m away, angle of rotation of reflected ray is
\( \alpha \approx \frac{y}{D} = \frac{0.002}{1} = 0.002 \, \text{radians} \).
Step 2: Relate to mirror rotation
If mirror rotates by \( \theta \), reflected ray rotates by \( 2\theta \).
So, \( 2\theta = \alpha \) ⇒ \( \theta = \frac{\alpha}{2} = \frac{0.002}{2} = 0.001 \, \text{radians} \).
Answer: Deflection of the coil = \( 0.001 \, \text{rad} \) or \( \frac{180}{\pi} \times 0.001 \approx 0.0573^\circ \).
Question 9.38
Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Answer & Explanation:
Concept: When needle's image coincides with itself, the needle is at the focal point of the system (lens + liquid layer + mirror). The system acts like a combination whose effective focal length equals the distance of needle.
Given: Refractive index of lens \( n_g = 1.50 \).
With liquid: Focal length of system, \( F_1 = 45.0 \, \text{cm} \).
Without liquid (lens in direct contact with mirror): Focal length, \( F_2 = 30.0 \, \text{cm} \).
Step 1: Focal length of lens alone
When lens is placed on plane mirror, needle at focus of lens-mirror combination.
For a lens on plane mirror, effective focal length \( F = \frac{f_{\text{lens}}}{2} \), where \( f_{\text{lens}} \) is focal length of lens alone.
So, \( F_2 = 30.0 = \frac{f}{2} \) ⇒ \( f = 60.0 \, \text{cm} \).
Step 2: With liquid layer
Now liquid between lens and mirror forms a plano-concave lens of refractive index \( n_l \) (unknown).
Combination: convex lens + liquid lens + mirror.
Equivalent focal length of lens-liquid combination before mirror: Let \( f_l \) be focal length of liquid lens (plano-concave).
For two lenses in contact: \( \frac{1}{f'} = \frac{1}{f} + \frac{1}{f_l} \), where \( f' \) is focal length of combination.
Then this combination on mirror gives effective focal length \( F_1 = \frac{f'}{2} \).
So, \( f' = 2F_1 = 90.0 \, \text{cm} \).
Step 3: Find \( f_l \)
\( \frac{1}{f'} = \frac{1}{f} + \frac{1}{f_l} \) ⇒ \( \frac{1}{90} = \frac{1}{60} + \frac{1}{f_l} \).
Solve for \( f_l \): \( \frac{1}{f_l} = \frac{1}{90} - \frac{1}{60} = \frac{2 - 3}{180} = -\frac{1}{180} \).
So, \( f_l = -180 \, \text{cm} \).
Step 4: Find \( n_l \) from lens maker's formula for liquid lens
Liquid lens: plano-concave, one surface plane (R1 = ∞), other surface concave with same radius as convex lens surface (say R).
For equiconvex lens: \( \frac{1}{f} = (n_g - 1) \left( \frac{2}{R} \right) \) ⇒ \( \frac{1}{60} = (1.5 - 1) \left( \frac{2}{R} \right) = 1 \times \frac{2}{R} \) wait: (1.5-1)=0.5, so \( \frac{1}{60} = 0.5 \times \frac{2}{R} = \frac{1}{R} \).
So, \( R = 60 \, \text{cm} \).
For plano-concave liquid lens: \( \frac{1}{f_l} = (n_l - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = (n_l - 1) \left( -\frac{1}{R} \right) \).
\( \frac{1}{-180} = (n_l - 1) \left( -\frac{1}{60} \right) \).
Cancel negatives: \( \frac{1}{180} = (n_l - 1) \frac{1}{60} \).
So, \( n_l - 1 = \frac{60}{180} = \frac{1}{3} \).
\( n_l = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333 \).
Given: Refractive index of lens \( n_g = 1.50 \).
With liquid: Focal length of system, \( F_1 = 45.0 \, \text{cm} \).
Without liquid (lens in direct contact with mirror): Focal length, \( F_2 = 30.0 \, \text{cm} \).
Step 1: Focal length of lens alone
When lens is placed on plane mirror, needle at focus of lens-mirror combination.
For a lens on plane mirror, effective focal length \( F = \frac{f_{\text{lens}}}{2} \), where \( f_{\text{lens}} \) is focal length of lens alone.
So, \( F_2 = 30.0 = \frac{f}{2} \) ⇒ \( f = 60.0 \, \text{cm} \).
Step 2: With liquid layer
Now liquid between lens and mirror forms a plano-concave lens of refractive index \( n_l \) (unknown).
Combination: convex lens + liquid lens + mirror.
Equivalent focal length of lens-liquid combination before mirror: Let \( f_l \) be focal length of liquid lens (plano-concave).
For two lenses in contact: \( \frac{1}{f'} = \frac{1}{f} + \frac{1}{f_l} \), where \( f' \) is focal length of combination.
Then this combination on mirror gives effective focal length \( F_1 = \frac{f'}{2} \).
So, \( f' = 2F_1 = 90.0 \, \text{cm} \).
Step 3: Find \( f_l \)
\( \frac{1}{f'} = \frac{1}{f} + \frac{1}{f_l} \) ⇒ \( \frac{1}{90} = \frac{1}{60} + \frac{1}{f_l} \).
Solve for \( f_l \): \( \frac{1}{f_l} = \frac{1}{90} - \frac{1}{60} = \frac{2 - 3}{180} = -\frac{1}{180} \).
So, \( f_l = -180 \, \text{cm} \).
Step 4: Find \( n_l \) from lens maker's formula for liquid lens
Liquid lens: plano-concave, one surface plane (R1 = ∞), other surface concave with same radius as convex lens surface (say R).
For equiconvex lens: \( \frac{1}{f} = (n_g - 1) \left( \frac{2}{R} \right) \) ⇒ \( \frac{1}{60} = (1.5 - 1) \left( \frac{2}{R} \right) = 1 \times \frac{2}{R} \) wait: (1.5-1)=0.5, so \( \frac{1}{60} = 0.5 \times \frac{2}{R} = \frac{1}{R} \).
So, \( R = 60 \, \text{cm} \).
For plano-concave liquid lens: \( \frac{1}{f_l} = (n_l - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = (n_l - 1) \left( -\frac{1}{R} \right) \).
\( \frac{1}{-180} = (n_l - 1) \left( -\frac{1}{60} \right) \).
Cancel negatives: \( \frac{1}{180} = (n_l - 1) \frac{1}{60} \).
So, \( n_l - 1 = \frac{60}{180} = \frac{1}{3} \).
\( n_l = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333 \).
📘 Exam Preparation Tip:
These exercise questions will help you understand light propagation and optical device principles. You'll learn to apply laws of reflection and refraction, solve problems using mirror and lens formulas, analyze combination of optical devices, understand working of microscopes, telescopes, and human eye, and calculate magnifying power and resolving power. Essential for optics and instrument design.