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Wave Optics
Physics XII - Chapter 10: Complete NCERT Exercise Solutions
Master Wave Optics with NCERT solutions for interference, diffraction, polarization, and Huygens' principle.
Question 10.1
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
Answer & Explanation:
(a) For reflected light:
• Reflection does not change medium, so speed remains \( c = 3 \times 10^8 \, \text{m/s} \).
• Frequency remains unchanged: \( \nu = \frac{c}{\lambda_{\text{air}}} \).
• Wavelength remains 589 nm.
(b) For refracted light (in water):
• Frequency remains same as in air: \( \nu \).
• Speed in water: \( v = \frac{c}{n} = \frac{3 \times 10^8}{1.33} \, \text{m/s} \).
• Wavelength in water: \( \lambda_{\text{water}} = \frac{\lambda_{\text{air}}}{n} = \frac{589}{1.33} \, \text{nm} \).
• Reflection does not change medium, so speed remains \( c = 3 \times 10^8 \, \text{m/s} \).
• Frequency remains unchanged: \( \nu = \frac{c}{\lambda_{\text{air}}} \).
• Wavelength remains 589 nm.
(b) For refracted light (in water):
• Frequency remains same as in air: \( \nu \).
• Speed in water: \( v = \frac{c}{n} = \frac{3 \times 10^8}{1.33} \, \text{m/s} \).
• Wavelength in water: \( \lambda_{\text{water}} = \frac{\lambda_{\text{air}}}{n} = \frac{589}{1.33} \, \text{nm} \).
Question 10.2
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Answer & Explanation:
(a) Light diverging from a point source:
Spherical wavefronts centered at the point source.
(b) Light emerging from convex lens with point source at focus:
Plane wavefronts (parallel beam).
(c) Wavefront from distant star intercepted by Earth:
Plane wavefronts (since star is very far, radius of curvature is extremely large).
Spherical wavefronts centered at the point source.
(b) Light emerging from convex lens with point source at focus:
Plane wavefronts (parallel beam).
(c) Wavefront from distant star intercepted by Earth:
Plane wavefronts (since star is very far, radius of curvature is extremely large).
Question 10.3
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is \( 3.0 \times 10^8 \, \text{m} \, \text{s}^{-1} \))
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Answer & Explanation:
(a) Speed in glass:
\( v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8 \, \text{m/s} \).
(b) Dependence on colour:
No, speed depends on wavelength (colour) due to dispersion.
Violet light has shorter wavelength and higher refractive index in glass than red light.
Since \( v = c/n \), violet travels slower than red in a glass prism.
\( v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8 \, \text{m/s} \).
(b) Dependence on colour:
No, speed depends on wavelength (colour) due to dispersion.
Violet light has shorter wavelength and higher refractive index in glass than red light.
Since \( v = c/n \), violet travels slower than red in a glass prism.
Question 10.4
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Answer & Explanation:
Step 1: Formula for bright fringe position
For nth bright fringe: \( y_n = \frac{n \lambda D}{d} \), where n = 0,1,2,...
Here, \( y_4 = 1.2 \, \text{cm} = 0.012 \, \text{m} \), \( n = 4 \), \( d = 0.28 \, \text{mm} = 2.8 \times 10^{-4} \, \text{m} \), \( D = 1.4 \, \text{m} \).
Step 2: Solve for \( \lambda \)
\( \lambda = \frac{y_n \cdot d}{n D} = \frac{0.012 \times 2.8 \times 10^{-4}}{4 \times 1.4} \, \text{m} \).
Calculate to get wavelength in meters, then convert to nm.
For nth bright fringe: \( y_n = \frac{n \lambda D}{d} \), where n = 0,1,2,...
Here, \( y_4 = 1.2 \, \text{cm} = 0.012 \, \text{m} \), \( n = 4 \), \( d = 0.28 \, \text{mm} = 2.8 \times 10^{-4} \, \text{m} \), \( D = 1.4 \, \text{m} \).
Step 2: Solve for \( \lambda \)
\( \lambda = \frac{y_n \cdot d}{n D} = \frac{0.012 \times 2.8 \times 10^{-4}}{4 \times 1.4} \, \text{m} \).
Calculate to get wavelength in meters, then convert to nm.
Question 10.5
In Young’s double-slit experiment using monochromatic light of wavelength \(\lambda\), the intensity of light at a point on the screen where path difference is \(\lambda\), is K units. What is the intensity of light at a point where path difference is \(\lambda/3\)?
Answer & Explanation:
Step 1: Relate intensity to phase difference
Intensity \( I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \), where \( \phi \) is phase difference.
Path difference \( \Delta x \) and phase difference: \( \phi = \frac{2\pi}{\lambda} \Delta x \).
Step 2: Case 1: \( \Delta x = \lambda \)
\( \phi = 2\pi \), \( \cos^2(\pi) = 1 \), so \( I_{\lambda} = 4I_0 = K \).
Step 3: Case 2: \( \Delta x = \lambda/3 \)
\( \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \).
\( \cos^2\left(\frac{\phi}{2}\right) = \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
So, \( I_{\lambda/3} = 4I_0 \times \frac{1}{4} = I_0 = \frac{K}{4} \).
Intensity \( I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \), where \( \phi \) is phase difference.
Path difference \( \Delta x \) and phase difference: \( \phi = \frac{2\pi}{\lambda} \Delta x \).
Step 2: Case 1: \( \Delta x = \lambda \)
\( \phi = 2\pi \), \( \cos^2(\pi) = 1 \), so \( I_{\lambda} = 4I_0 = K \).
Step 3: Case 2: \( \Delta x = \lambda/3 \)
\( \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \).
\( \cos^2\left(\frac{\phi}{2}\right) = \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
So, \( I_{\lambda/3} = 4I_0 \times \frac{1}{4} = I_0 = \frac{K}{4} \).
Question 10.6
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer & Explanation:
Given: \( \lambda_1 = 650 \, \text{nm} \), \( \lambda_2 = 520 \, \text{nm} \). Assume \( d \) and \( D \) given or cancel out.
(a) Position of 3rd bright fringe for \( \lambda_1 \):
\( y_3^{(1)} = \frac{3 \lambda_1 D}{d} \). Need \( D/d \) ratio if not given.
(b) Coincidence of bright fringes:
Let nth fringe of \( \lambda_1 \) coincide with mth fringe of \( \lambda_2 \):
\( \frac{n \lambda_1 D}{d} = \frac{m \lambda_2 D}{d} \) ⇒ \( \frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5} \).
Smallest integers: \( n = 4 \), \( m = 5 \).
Least distance: \( y = \frac{4 \lambda_1 D}{d} = \frac{5 \lambda_2 D}{d} \).
(a) Position of 3rd bright fringe for \( \lambda_1 \):
\( y_3^{(1)} = \frac{3 \lambda_1 D}{d} \). Need \( D/d \) ratio if not given.
(b) Coincidence of bright fringes:
Let nth fringe of \( \lambda_1 \) coincide with mth fringe of \( \lambda_2 \):
\( \frac{n \lambda_1 D}{d} = \frac{m \lambda_2 D}{d} \) ⇒ \( \frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5} \).
Smallest integers: \( n = 4 \), \( m = 5 \).
Least distance: \( y = \frac{4 \lambda_1 D}{d} = \frac{5 \lambda_2 D}{d} \).
Question 10.7
In a double-slit experiment the angular width of a fringe is found to be \( 0.2^\circ \) on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Answer & Explanation:
Step 1: Angular width formula
Angular width \( \theta = \frac{\lambda}{d} \) (in radians).
Given \( \theta_{\text{air}} = 0.2^\circ = 0.2 \times \frac{\pi}{180} \, \text{rad} \).
Step 2: Effect of immersion in water
Wavelength in water: \( \lambda_w = \frac{\lambda}{n} \).
Angular width in water: \( \theta_w = \frac{\lambda_w}{d} = \frac{\lambda}{n d} = \frac{\theta_{\text{air}}}{n} \).
\( \theta_w = \frac{0.2^\circ}{(4/3)} = 0.15^\circ \).
Angular width \( \theta = \frac{\lambda}{d} \) (in radians).
Given \( \theta_{\text{air}} = 0.2^\circ = 0.2 \times \frac{\pi}{180} \, \text{rad} \).
Step 2: Effect of immersion in water
Wavelength in water: \( \lambda_w = \frac{\lambda}{n} \).
Angular width in water: \( \theta_w = \frac{\lambda_w}{d} = \frac{\lambda}{n d} = \frac{\theta_{\text{air}}}{n} \).
\( \theta_w = \frac{0.2^\circ}{(4/3)} = 0.15^\circ \).
Question 10.8
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer & Explanation:
Brewster's law: \( \tan i_p = n \), where \( n \) is refractive index of glass relative to air.
\( i_p = \tan^{-1}(n) = \tan^{-1}(1.5) \).
Calculate: \( i_p \approx 56.31^\circ \).
\( i_p = \tan^{-1}(n) = \tan^{-1}(1.5) \).
Calculate: \( i_p \approx 56.31^\circ \).
Question 10.9
Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Answer & Explanation:
Part 1: Wavelength and frequency
Reflection does not change medium, so:
• Wavelength: 5000 Å = 500 nm.
• Frequency: \( \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{500 \times 10^{-9}} \, \text{Hz} \).
Part 2: Angle for reflected ray normal to incident ray
Reflected ray normal to incident ray means angle between them = 90°.
If angle of incidence = \( i \), angle of reflection = \( i \).
Then \( i + i = 90^\circ \) ⇒ \( i = 45^\circ \).
Reflection does not change medium, so:
• Wavelength: 5000 Å = 500 nm.
• Frequency: \( \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{500 \times 10^{-9}} \, \text{Hz} \).
Part 2: Angle for reflected ray normal to incident ray
Reflected ray normal to incident ray means angle between them = 90°.
If angle of incidence = \( i \), angle of reflection = \( i \).
Then \( i + i = 90^\circ \) ⇒ \( i = 45^\circ \).
Question 10.10
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer & Explanation:
Fresnel distance formula: \( Z_F = \frac{a^2}{\lambda} \), beyond which diffraction effects become significant.
Here, aperture \( a = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \), \( \lambda = 400 \, \text{nm} = 4 \times 10^{-7} \, \text{m} \).
\( Z_F = \frac{(4 \times 10^{-3})^2}{4 \times 10^{-7}} = \frac{16 \times 10^{-6}}{4 \times 10^{-7}} = 40 \, \text{m} \).
So, for distances much less than 40 m, ray optics is a good approximation.
Here, aperture \( a = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \), \( \lambda = 400 \, \text{nm} = 4 \times 10^{-7} \, \text{m} \).
\( Z_F = \frac{(4 \times 10^{-3})^2}{4 \times 10^{-7}} = \frac{16 \times 10^{-6}}{4 \times 10^{-7}} = 40 \, \text{m} \).
So, for distances much less than 40 m, ray optics is a good approximation.
Question 10.11
The 6563 Å H\(\alpha\) line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
Answer & Explanation:
Doppler formula for light: \( \frac{\Delta \lambda}{\lambda} \approx \frac{v}{c} \) for \( v \ll c \).
Here, \( \lambda = 6563 \, \text{Å} \), \( \Delta \lambda = 15 \, \text{Å} \) (redshift ⇒ receding).
\( \frac{v}{c} = \frac{15}{6563} \).
\( v = \frac{15}{6563} \times 3 \times 10^8 \, \text{m/s} \).
Calculate to get speed in m/s or km/s.
Here, \( \lambda = 6563 \, \text{Å} \), \( \Delta \lambda = 15 \, \text{Å} \) (redshift ⇒ receding).
\( \frac{v}{c} = \frac{15}{6563} \).
\( v = \frac{15}{6563} \times 3 \times 10^8 \, \text{m/s} \).
Calculate to get speed in m/s or km/s.
Question 10.12
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
Answer & Explanation:
Corpuscular theory (Newton): Light consists of particles. When entering denser medium (water), particles are attracted by the medium, increasing speed.
Experimental fact: Speed of light in water is less than in vacuum (\( c/n \), n>1).
Consistent theory: Wave theory (Huygens) predicts \( v = c/n \), which matches experiment. Later, electromagnetic theory confirmed this.
Experimental fact: Speed of light in water is less than in vacuum (\( c/n \), n>1).
Consistent theory: Wave theory (Huygens) predicts \( v = c/n \), which matches experiment. Later, electromagnetic theory confirmed this.
Question 10.13
You have learnt in the text how Huygens' principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Answer & Explanation:
Using Huygens' construction:
1. Consider point object O at distance u from mirror.
2. Each point on wavefront from O acts as source of secondary wavelet.
3. After reflection, shape of wavefront is spherical, appearing to come from point I behind mirror.
4. Geometry shows I is at distance u behind mirror, same as object distance in front.
5. Since rays appear to diverge from I (no actual light behind mirror), image is virtual.
1. Consider point object O at distance u from mirror.
2. Each point on wavefront from O acts as source of secondary wavelet.
3. After reflection, shape of wavefront is spherical, appearing to come from point I behind mirror.
4. Geometry shows I is at distance u behind mirror, same as object distance in front.
5. Since rays appear to diverge from I (no actual light behind mirror), image is virtual.
Question 10.14
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) direction of propagation.
(iii) motion of the source and/or observer.
(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend?
(i) nature of the source.
(ii) direction of propagation.
(iii) motion of the source and/or observer.
(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend?
Answer & Explanation:
(a) Speed of light in vacuum:
Independent of (i), (ii), (iii), (iv), (v). It is a universal constant \( c \).
(b) Speed of light in a medium:
• Depends on (iv) wavelength (due to dispersion).
• May depend on (v) intensity for very high intensities (non-linear optics).
• Independent of (i), (ii), (iii) for normal conditions.
Independent of (i), (ii), (iii), (iv), (v). It is a universal constant \( c \).
(b) Speed of light in a medium:
• Depends on (iv) wavelength (due to dispersion).
• May depend on (v) intensity for very high intensities (non-linear optics).
• Independent of (i), (ii), (iii) for normal conditions.
Question 10.15
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
Answer & Explanation:
For light in vacuum:
According to special relativity, speed of light is constant in all inertial frames. Only relative motion between source and observer matters. Hence formulas are identical.
For light in a medium:
Speed of light is \( c/n \) relative to the medium. Motion of source or observer relative to medium matters. Therefore, formulas for the two cases (source moving vs observer moving) would differ, similar to sound.
According to special relativity, speed of light is constant in all inertial frames. Only relative motion between source and observer matters. Hence formulas are identical.
For light in a medium:
Speed of light is \( c/n \) relative to the medium. Motion of source or observer relative to medium matters. Therefore, formulas for the two cases (source moving vs observer moving) would differ, similar to sound.
Question 10.16
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?
Answer & Explanation:
Angular width of fringe: \( \theta = \frac{\lambda}{d} \).
Given \( \theta = 0.1^\circ = 0.1 \times \frac{\pi}{180} \, \text{rad} \), \( \lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \).
\( d = \frac{\lambda}{\theta} = \frac{6 \times 10^{-7}}{0.1 \times (\pi/180)} \, \text{m} \).
Calculate to get slit separation in meters or mm.
Given \( \theta = 0.1^\circ = 0.1 \times \frac{\pi}{180} \, \text{rad} \), \( \lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \).
\( d = \frac{\lambda}{\theta} = \frac{6 \times 10^{-7}}{0.1 \times (\pi/180)} \, \text{m} \).
Calculate to get slit separation in meters or mm.
Question 10.17
Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Answer & Explanation:
(a) Width of central band is inversely proportional to slit width. Doubling slit width halves angular width. Intensity increases (more light passes).
(b) Diffraction from each slit determines the envelope of intensity, within which interference fringes are modulated.
(c) Due to Fresnel diffraction, waves diffracted from edges of obstacle interfere constructively at center, producing Poisson/Arago spot.
(d) Wavelength of sound (~m) comparable to obstacle size, so significant diffraction. Wavelength of light (~500 nm) much smaller, negligible diffraction over such scale.
(e) For apertures >> wavelength, diffraction is negligible. Optical instruments have sizes much larger than λ, so ray optics is excellent approximation for image location, though not for fine effects.
(b) Diffraction from each slit determines the envelope of intensity, within which interference fringes are modulated.
(c) Due to Fresnel diffraction, waves diffracted from edges of obstacle interfere constructively at center, producing Poisson/Arago spot.
(d) Wavelength of sound (~m) comparable to obstacle size, so significant diffraction. Wavelength of light (~500 nm) much smaller, negligible diffraction over such scale.
(e) For apertures >> wavelength, diffraction is negligible. Optical instruments have sizes much larger than λ, so ray optics is excellent approximation for image location, though not for fine effects.
Question 10.18
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Answer & Explanation:
Concept: For negligible diffraction, size of obstacle/aperture >> λ. Here obstacle hill of height 50 m halfway.
Use Fresnel zone formula: For waves to pass over hill with little diffraction, path difference ≤ λ/4.
Geometry: Path difference ≈ \( \frac{h^2}{2d} \) where h = 50 m, d = half distance = 20 km = 20000 m.
Set \( \frac{h^2}{2d} \leq \frac{\lambda}{4} \).
Solve for λ: \( \lambda \geq \frac{2h^2}{d} \).
Calculate to get minimum λ for significant diffraction; question asks longest λ without diffraction, so use equality.
Use Fresnel zone formula: For waves to pass over hill with little diffraction, path difference ≤ λ/4.
Geometry: Path difference ≈ \( \frac{h^2}{2d} \) where h = 50 m, d = half distance = 20 km = 20000 m.
Set \( \frac{h^2}{2d} \leq \frac{\lambda}{4} \).
Solve for λ: \( \lambda \geq \frac{2h^2}{d} \).
Calculate to get minimum λ for significant diffraction; question asks longest λ without diffraction, so use equality.
Question 10.19
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Answer & Explanation:
Single‑slit minima condition: \( a \sin \theta = n \lambda \), n = 1,2,...
For small θ, \( \sin \theta \approx \tan \theta = \frac{y}{D} \).
Here, first minimum n=1, y = 2.5 mm = 2.5 × 10⁻³ m, D = 1 m, λ = 500 × 10⁻⁹ m.
\( a \cdot \frac{y}{D} = \lambda \) ⇒ \( a = \frac{\lambda D}{y} \).
Calculate a in meters.
For small θ, \( \sin \theta \approx \tan \theta = \frac{y}{D} \).
Here, first minimum n=1, y = 2.5 mm = 2.5 × 10⁻³ m, D = 1 m, λ = 500 × 10⁻⁹ m.
\( a \cdot \frac{y}{D} = \lambda \) ⇒ \( a = \frac{\lambda D}{y} \).
Calculate a in meters.
Question 10.20
Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
Answer & Explanation:
(a) Aircraft reflects TV signals causing multipath interference. Slight changes in path difference due to aircraft motion cause fluctuating interference at receiver, leading to shaking picture.
(b) Linear superposition holds when wave equation is linear. For light in most media, electric and magnetic fields obey linear Maxwell’s equations. Hence displacements add linearly.
(b) Linear superposition holds when wave equation is linear. For light in most media, electric and magnetic fields obey linear Maxwell’s equations. Hence displacements add linearly.
Question 10.21
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of \( n\lambda/a \). Justify this by suitably dividing the slit to bring out the cancellation.
Answer & Explanation:
Divide slit into two halves for first minimum (n=1).
Path difference from top edge and midpoint = \( \frac{a}{2} \sin \theta \).
For \( \sin \theta = \lambda/a \), this path difference = λ/2 → phase difference = π.
Waves from corresponding points in two halves cancel pairwise.
For n=2, divide into four parts; each quarter cancels with opposite quarter, etc.
General: divide slit into 2n equal parts; path difference between adjacent parts = λ/2 → cancellation.
Path difference from top edge and midpoint = \( \frac{a}{2} \sin \theta \).
For \( \sin \theta = \lambda/a \), this path difference = λ/2 → phase difference = π.
Waves from corresponding points in two halves cancel pairwise.
For n=2, divide into four parts; each quarter cancels with opposite quarter, etc.
General: divide slit into 2n equal parts; path difference between adjacent parts = λ/2 → cancellation.
📘 Exam Preparation Tip:
These exercise questions will help you understand wave nature of light and interference phenomena. You'll learn to analyze Young's double slit experiment, calculate fringe width and intensity, understand diffraction patterns (single slit and diffraction grating), apply Huygens' principle to wavefronts, and analyze polarization of light. Crucial for understanding optical phenomena and advanced optics.