Chapter 10: Light Reflection and Refraction
Solved NCERT Solutions for Light Reflection and Refraction. Step-by-step answers for mirror and lens formulas for Class 10.
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Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Explanation:
A lens must be transparent to allow light to pass through and have a uniform refractive index to bend light predictably.
Analysis of options:
• Water: Can form lenses (water droplets act as convex lenses)
• Glass: Traditional lens material (transparent, moldable)
• Plastic: Modern lens material (acrylic, polycarbonate lenses)
• Clay: Opaque material → cannot transmit light → cannot form image
Key requirement: Transparency + ability to refract light. Clay is opaque and scatters light.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Explanation:
Concave mirror image characteristics based on object position:
| Object Position | Image Characteristics |
|---|---|
| Beyond C | Real, inverted, diminished (between F and C) |
| At C | Real, inverted, same size (at C) |
| Between C and F | Real, inverted, enlarged (beyond C) |
| At F | Real, inverted, highly enlarged (at infinity) |
| Between P and F | Virtual, erect, enlarged (behind mirror) |
Why this happens:
When object is between pole (P) and focus (F), reflected rays diverge. Their backward extensions meet behind the mirror, forming a virtual, erect, and magnified image (as in shaving/makeup mirrors).
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Explanation:
For a convex lens:
| Object Position | Image Position | Image Size |
|---|---|---|
| At infinity | At F₂ | Highly diminished |
| Beyond 2F₁ | Between F₂ and 2F₂ | Diminished |
| At 2F₁ | At 2F₂ | Same size |
| Between F₁ and 2F₁ | Beyond 2F₂ | Enlarged |
| At F₁ | At infinity | Highly enlarged |
| Between F₁ and O | Same side as object | Virtual, enlarged |
Mathematical proof using lens formula:
For same size: magnification m = -1 (negative for real image)
m = v/u = -1 ⇒ v = -u
Using lens formula: 1/v - 1/u = 1/f
1/(-u) - 1/u = 1/f ⇒ -2/u = 1/f ⇒ u = -2f
So object at 2f gives image at 2f with same size.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Explanation:
Sign convention (New Cartesian):
• Distances measured in direction of incident light = positive
• Opposite direction = negative
Focal length signs:
• Concave mirror: Focus in front → negative f
• Convex mirror: Focus behind → positive f
• Convex lens: Focus on opposite side → positive f
• Concave lens: Focus on same side → negative f
Given: f = -15 cm for both
• Mirror with f negative → concave mirror
• Lens with f negative → concave lens
But option (a) says "both concave" - concave mirror ✓ but concave lens ✗
Wait, let's check: The question says "thin spherical lens" with f = -15 cm
Negative f for lens means concave (diverging) lens.
Actually, re-reading options:
(a) both concave → concave mirror + concave lens ✓
(b) both convex → both would have positive f ✗
(c) concave mirror + convex lens → f(concave) negative, f(convex) positive ✗
(d) convex mirror + concave lens → f(convex) positive, f(concave) negative ✗
So (a) is correct: concave mirror (f negative) and concave lens (f negative).
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.
Explanation:
Image characteristics of different mirrors:
| Mirror Type | Image Nature | Erect for All Positions? |
|---|---|---|
| Plane Mirror | Always virtual, erect, same size | ✓ Yes |
| Concave Mirror | Can be real/inverted or virtual/erect depending on object position | ✗ No (only erect when object between P and F) |
| Convex Mirror | Always virtual, erect, diminished | ✓ Yes |
Key points:
• Plane mirrors: Always give erect image regardless of distance
• Convex mirrors: Always give erect image (used in rear-view mirrors)
• Concave mirrors: Give erect image ONLY when object is between pole and focus (close to mirror); otherwise inverted
Since the statement says "no matter how far you stand" (all distances), both plane and convex qualify.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Explanation:
For reading small letters (magnification):
• Need a magnifying glass → convex lens (converging)
• Concave lens always gives diminished image → unsuitable
Why shorter focal length (5 cm) is better than 50 cm:
• Magnification m = 1 + D/f (where D = least distance of distinct vision ≈ 25 cm)
• For f = 5 cm: m = 1 + 25/5 = 6× magnification
• For f = 50 cm: m = 1 + 25/50 = 1.5× magnification
• Shorter focal length → greater magnification
Practical consideration: A 5 cm focal length convex lens held close to text acts as a simple microscope, producing virtual, erect, and enlarged image suitable for reading small print.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
1. Range of object distance:
For a concave mirror to produce an erect image, the object must be placed between the pole (P) and the principal focus (F).
Given f = -15 cm (negative for concave mirror)
So object distance: 0 < u < 15 cm (between mirror and focus)
2. Nature of image:
• Virtual (cannot be projected on screen)
• Erect (upright)
• Behind the mirror
3. Size comparison:
The image is larger than the object (magnified).
Magnification m = -v/u > 1 (since v is positive for virtual image in sign convention)
4. Ray diagram:
1. Concave mirror with principal axis, pole P, focus F, center C
2. Object AB placed between P and F (close to mirror)
3. Ray 1: Parallel to axis → reflects through F
4. Ray 2: Through F → reflects parallel to axis
5. Both reflected rays diverge; extend backward to meet at A'B'
6. Image A'B' is virtual, erect, enlarged, behind mirror
Verification using mirror formula:
For u = 10 cm (between 0 and 15 cm):
1/v + 1/u = 1/f
1/v + 1/(-10) = 1/(-15)
1/v = -1/15 + 1/10 = (-2 + 3)/30 = 1/30
v = +30 cm (positive = virtual image behind mirror)
m = -v/u = -30/(-10) = +3 (erect, 3× magnified)
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
| Situation | Mirror Type | Reason |
|---|---|---|
| (a) Headlights of a car | Concave Mirror | • Produces parallel beam of light when bulb is at focus • Converges light forward for powerful illumination • Creates concentrated beam for long-distance visibility |
| (b) Side/rear-view mirror of a vehicle | Convex Mirror | • Always gives diminished, erect, virtual image • Wider field of view (sees more area) • Allows driver to see traffic behind without turning head • Reduces blind spots |
| (c) Solar furnace | Concave Mirror | • Concentrates sunlight at focus where high temperature is produced • Large aperture collects maximum solar energy • Parabolic shape focuses parallel rays to single point • Can generate temperatures up to 3000°C |
Memory aid: Concave = Concentrate (light/heat); Convex = View (wide view).
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Yes, the lens will produce a complete image, but it will be less bright.
Experimental verification:
1. Take a convex lens and project image of a distant object on screen.
2. Note the clear, complete image formed.
3. Cover half the lens with black paper.
4. Observe the image again on screen.
Observations:
• Image is still complete (shows entire object)
• Image is less bright (dimmer)
• Sharpness may slightly reduce if covered part includes center
Explanation:
• Light from each point of object passes through all parts of lens to form image.
• When half is covered, light from object points still passes through uncovered half.
• Each point on object still has rays reaching corresponding point on image.
• However, intensity reduces because less light reaches screen (half the light blocked).
Analogy: Like reducing aperture in camera – smaller opening → less light → dimmer image but still complete.
Exception: If covered part includes optical center, image may be slightly blurred as central rays help in proper focusing.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Given:
• Object height (h) = +5 cm
• Object distance (u) = -25 cm (negative by convention)
• Focal length (f) = +10 cm (positive for converging/convex lens)
Step 1: Find image distance (v) using lens formula:
1/v = 1/f + 1/u
1/v = 1/10 + 1/(-25)
1/v = 1/10 - 1/25 = (5 - 2)/50 = 3/50
v = +50/3 ≈ +16.67 cm
Step 2: Find image height (h') using magnification:
h' = h × (v/u)
h' = 5 × (16.67/(-25))
h' = 5 × (-0.6668) ≈ -3.33 cm
Step 3: Determine nature:
• Position: 16.67 cm from lens on opposite side
• Size: 3.33 cm (diminished: 2/3 of object)
• Nature: Real, inverted, diminished
Step 4: Ray diagram:
1. Convex lens with principal axis, optical center O, foci F₁ and F₂
2. Object AB = 5 cm at 25 cm left of lens (beyond 2F₁)
3. Ray 1: Parallel to axis → refracts through F₂
4. Ray 2: Through optical center O → goes straight
5. Ray 3: Through F₁ → refracts parallel to axis
6. All rays meet at A'B' between F₂ and 2F₂
7. Image A'B' is real, inverted, diminished
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Given:
• Focal length (f) = -15 cm (negative for concave lens)
• Image distance (v) = -10 cm (negative for virtual image formed by concave lens)
• Object distance (u) = ?
Using lens formula:
-1/u = 1/f - 1/v
-1/u = 1/(-15) - 1/(-10)
-1/u = -1/15 + 1/10
-1/u = (-2 + 3)/30 = 1/30
1/u = -1/30
u = -30 cm
Result: Object is placed 30 cm from the concave lens.
Ray diagram:
1. Concave lens with principal axis, optical center O, focus F₁ on left
2. Object AB at 30 cm left of lens
3. Ray 1: Parallel to axis → refracts as if coming from F₁
4. Ray 2: Towards optical center O → goes straight
5. Both refracted rays diverge; extend backward to meet at A'B'
6. Image A'B' is virtual, erect, diminished, at 10 cm from lens on same side as object
Verification:
For concave lens, image is always virtual, erect, diminished, and between focus and lens regardless of object position.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Given:
• Object distance (u) = -10 cm (negative by convention)
• Focal length (f) = +15 cm (positive for convex mirror)
• Image distance (v) = ?
Using mirror formula:
1/v = 1/f - 1/u
1/v = 1/15 - 1/(-10)
1/v = 1/15 + 1/10
1/v = (2 + 3)/30 = 5/30 = 1/6
v = +6 cm
Magnification:
|m| = 0.6 < 1 → Diminished image.
Result:
• Position: 6 cm behind the mirror
• Nature: Virtual, erect, diminished
Note: This matches the general property of convex mirrors: always produce virtual, erect, diminished images regardless of object position.
The magnification produced by a plane mirror is +1. What does this mean?
Magnification (m) = +1 means:
1. Size information:
• |m| = 1 → Image size = Object size (neither enlarged nor diminished)
• Height of image = Height of object
2. Orientation information:
• m positive (+1) → Image is erect (upright), not inverted
• Same orientation as object
3. Additional implications for plane mirrors:
• Virtual image (cannot be projected on screen)
• Image distance = Object distance (as far behind mirror as object is in front)
• Laterally inverted (left-right reversal)
Mathematical representation:
Therefore:
h' = h (same size)
-v/u = 1 ⇒ v = -u (image distance = object distance with sign change)
• Your image is 2 m behind the mirror
• Image is exactly your height
• Image faces you (erect) but your right hand appears as left hand in image
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given:
• Object height (h) = +5.0 cm
• Object distance (u) = -20 cm
• Radius of curvature (R) = +30 cm (positive for convex mirror)
• Focal length f = R/2 = +15 cm
Step 1: Find image distance (v) using mirror formula:
1/v = 1/f - 1/u
1/v = 1/15 - 1/(-20)
1/v = 1/15 + 1/20 = (4 + 3)/60 = 7/60
v = +60/7 ≈ +8.57 cm
Step 2: Find magnification (m):
|m| < 1 → Diminished image.
Step 3: Find image height (h'):
h' = m × h = (3/7) × 5 = 15/7 ≈ 2.14 cm
• Position: 8.57 cm behind the mirror
• Nature: Virtual, erect, diminished
• Size: 2.14 cm (about 43% of object size)
Verification: Convex mirrors always produce virtual, erect, diminished images regardless of object position.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Given:
• Object height (h) = +7.0 cm
• Object distance (u) = -27 cm
• Focal length (f) = -18 cm (negative for concave mirror)
Step 1: Find image distance (v) using mirror formula:
1/v = 1/f - 1/u
1/v = 1/(-18) - 1/(-27)
1/v = -1/18 + 1/27
1/v = (-3 + 2)/54 = -1/54
v = -54 cm
Screen position: Screen should be placed 54 cm in front of the mirror.
Step 2: Find magnification (m):
|m| = 2 > 1 → Enlarged image (2×).
Step 3: Find image height (h'):
h' = m × h = -2 × 7 = -14 cm
Results:
• Screen position: 54 cm in front of mirror
• Image size: 14 cm (twice the object size)
• Nature: Real, inverted, enlarged
Verification of object position:
Object at u = -27 cm, f = -18 cm ⇒ Object between F and C (since 27 > 18 but < 36 if C at 2f = 36 cm)
According to mirror rules: Object between F and C → Image beyond C, real, inverted, enlarged ✓
Find the focal length of a lens of power - 2.0 D. What type of lens is this?
Given: Power (P) = -2.0 D
Relationship: Power P = 1/f (where f is in meters)
Type of lens:
• Negative power → Diverging lens
• Negative focal length → Concave lens
Result: It's a concave (diverging) lens with focal length 50 cm.
Additional information:
• Concave lenses always have negative power
• Used to correct myopia (nearsightedness)
• Always produce virtual, erect, diminished images
• Diopter (D) = 1/f(m), so f = 1/P in meters
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Given: Power (P) = +1.5 D
Focal length calculation:
f = 1/P = 1/(+1.5) = +0.666... m ≈ +0.67 m = +67 cm
Type of lens:
• Positive power → Converging lens
• Positive focal length → Convex lens
Result: It's a convex (converging) lens with focal length about 67 cm.
Medical application:
• +1.5 D convex lens is used to correct hypermetropia (farsightedness)
• Helps converge light rays onto retina for people with shorter eyeballs
• Typical reading glasses for presbyopia (age-related farsightedness) have powers from +1.0 D to +3.0 D
Comparison:
• Myopia correction: Negative power (concave lenses)
• Hypermetropia correction: Positive power (convex lenses)