Light Reflection & Refraction Test Class 10 — 40 Marks
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PERIODIC TEST — LIGHT · REFLECTION AND REFRACTION (2026–27) | CLASS 10 | SUB: SCIENCE
Max Marks: 40 | Time: 90 minutes (1 hour 30 minutes)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A · MCQ | 12 | 12 | 15 m | Quick recall |
| B · Short Answer | 8 | 4 | 20 m | Brief answers |
| C · Long Answer | 12 | 4 | 30 m | Detailed steps |
| D · Case Based | 8 | 2 | 20 m | Analysis |
| Review | — | — | 5 m | Cross-check |
SECTION A — Multiple Choice (12 × 1 = 12 Marks)
Choose the correct option for Q1–Q12. No internal choice.
Rays of light parallel to the principal axis of a concave mirror, after reflection, meet at the:
Marking Scheme: Correct option: 1M (No partial)
Answer: (a). Rays parallel to the principal axis of a concave mirror converge at the principal focus.
The radius of curvature of a spherical mirror is 20 cm. Its focal length is:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). f = R/2 = 20/2 = 10 cm.
The image formed by a plane mirror is always:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). Plane mirrors always give virtual, erect images of the same size as the object.
Light travels fastest in:
Marking Scheme: Correct option: 1M (No partial)
Answer: (c). Speed in vacuum is 3 × 10⁸ m/s, the highest possible speed of light.
The SI unit of power of a lens is the:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). Power of a lens is measured in dioptre; 1 D = 1 m⁻¹.
When a ray of light passes from air into water, it:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). Going from a rarer (air) to a denser (water) medium bends light towards the normal.
Absolute refractive index of a medium is defined as:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). n = c / v, i.e., speed of light in vacuum divided by speed in the medium.
A concave lens always forms an image that is:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). A concave (diverging) lens always forms a virtual, erect and diminished image.
A ray of light passing through the centre of curvature of a concave mirror:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). It hits the mirror normally (angle of incidence 0°), so it retraces its path.
Magnification of a mirror is defined as:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). m = height of image (h′) ÷ height of object (h).
For a real object, a concave mirror keeps a real image only when the object is:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). Only when the object lies beyond the focus (or at the focus) is the image real.
A convex mirror is used as a rear-view mirror in vehicles because it:
Marking Scheme: Correct option: 1M (No partial)
Answer: (b). The convex mirror gives erect and diminished images covering a wider area of the road.
SECTION B — Short Answer (4 × 2 = 8 Marks)
State the two laws of reflection of light.
Marking Scheme: Law 1: 1M · Law 2: 1M
Law of incidence
The angle of incidence equals the angle of reflection.
Law of coplanarity
The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
Define principal focus of a concave mirror. Where does the image form when an object is placed at the focus?
Marking Scheme: Focus definition: 1M · Image position: 1M
Definition
The point on the principal axis where rays parallel to the axis converge after reflection.
Object at focus
The image forms at infinity (rays emerge parallel).
The absolute refractive index of diamond is 2.42. Calculate the speed of light in diamond. (Speed of light in vacuum c = 3 × 10⁸ m/s.)
Marking Scheme: Formula: 1M · Calculation with unit: 1M
Formula
n = c / v, so v = c / n
Substitution
v = (3 × 10⁸) / 2.42
Result
v ≈ 1.24 × 10⁸ m/s
✅ Final Answer: 1.24 × 10⁸ m/s
Explain with a reason why a convex mirror is preferred as a rear-view mirror in vehicles.
— OR / अथवा
OR option: State the laws of refraction of light. What is the refractive index of a medium?
[Note: Attempt any ONE]
Marking Scheme: Reason: 1M · Property used: 1M
Option 1 — Convex rear-view mirror
Choice of mirror
A convex mirror is used as a rear-view mirror in vehicles.
Two reasons
It always produces an erect and diminished image, and it gives a much wider field of view so the driver can see more of the road behind.
Option 2 — Laws of refraction
First law (Snell’s law)
The ratio of the sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media.
Second law
The incident ray, the refracted ray and the normal all lie in the same plane.
Refractive index
It is the ratio of the speed of light in vacuum to its speed in the given medium.
SECTION C — Long Answer (4 × 3 = 12 Marks)
A concave mirror has a radius of curvature of 40 cm. An object of height 5 cm is placed 30 cm in front of it. Using the mirror formula, find the position, nature and size of the image.
Marking Scheme (3M Math): Formula 0.5M · Substitution 1M · Steps 1M · Answer+unit 0.5M
Given data
R = 40 cm ⇒ f = −20 cm, u = −30 cm (sign convention), h = 5 cm
Mirror formula
1/v + 1/u = 1/f ⇒ 1/v = 1/f − 1/u
Substitution
1/v = (−1/20) − (−1/30) = −3/60 + 2/60 = −1/60
Image distance
v = −60 cm (in front of the mirror, so real)
Magnification
m = −v/u = −(−60)/(−30) = −2 (inverted)
Image height
h′ = m × h = 2 × 5 = 10 cm
✅ Final Answer: Image 60 cm in front, real and inverted, 10 cm tall.
Define power of a lens and give its unit. A converging lens has a power of + 2.0 D. Find its focal length.
Marking Scheme (3M Math): Definition 1M · Formula 1M · Answer+unit 1M
Definition
Power of a lens is the reciprocal of its focal length in metres: P = 1/f.
Unit
Dioptre (D); 1 D = 1 m⁻¹.
Focal length
f = 1/P = 1/2.0 = 0.5 m = 50 cm
Nature
Positive power means a convex (converging) lens.
✅ Final Answer: f = 50 cm (convex lens)
A convex lens of focal length 10 cm forms an image of an object placed 15 cm away on the principal axis. Find the position and nature of the image by the lens formula.
— OR / अथवा
OR option: An object of height 6 cm is placed 20 cm in front of a concave lens of focal length 20 cm. Find the position, nature and size of the image.
[Note: Attempt any ONE]
Marking Scheme (3M Math): Formula 0.5M · Substitution 1M · Steps 1M · Answer+unit 0.5M
Option 1 — Convex lens, f = +10 cm
Given data
f = +10 cm, u = −15 cm
Lens formula
1/v − 1/u = 1/f ⇒ 1/v = 1/f + 1/u
Substitution
1/v = 1/10 + (−1/15) = 3/30 − 2/30 = 1/30
Position
v = +30 cm (on the other side of the lens, real)
Magnification
m = v/u = 30/(−15) = −2 (inverted and magnified)
Option 2 — Concave lens, f = −20 cm
Given data
f = −20 cm, u = −20 cm, h = 6 cm
Lens formula
1/v − 1/u = 1/f ⇒ 1/v = 1/f + 1/u
Substitution
1/v = (−1/20) + (−1/20) = −2/20 = −1/10
Position
v = −10 cm (same side as object, virtual)
Size
m = v/u = (−10)/(−20) = +1/2; h′ = (1/2) × 6 = 3 cm (erect, diminished)
✅ Option 1: 30 cm other side, real inverted 2×. ✅ Option 2: 10 cm same side, virtual erect, 3 cm.
Explain why the bottom of a pool or a tank filled with water appears raised. Draw a labelled ray diagram to support your answer.
— OR / अथवा
OR option: Distinguish between a real image and a virtual image by giving two points of difference.
[Note: Attempt any ONE]
Marking Scheme (3M Theory): Point1 1M · Point2 example/diagram 1M · Point3 conclusion 1M
Option 1 — Apparent depth
Refraction
Light from the bottom travels from water (denser) to air (rarer).
Bending
It bends away from the normal at the surface, so rays appear to come from a point higher than the real bottom.
Result
The bottom appears raised to a shallow apparent depth.
Diagram
Schematic — not to scale: a vertical ray from the bottom bends away from the normal at the water surface; the eye extends the emerging ray backward to a higher apparent position.
Option 2 — Real vs virtual image
Point 1 — Rays
A real image is formed where the reflected/refracted rays actually meet; a virtual image forms where the extended rays appear to meet.
Point 2 — Reception
A real image can be obtained on a screen; a virtual image cannot be obtained on a screen.
Examples
A concave mirror of an object beyond focus gives a real image; a plane mirror always gives a virtual image.
SECTION D — Case Based (2 × 4 = 8 Marks)
Case study. A student fixes a concave mirror of focal length 20 cm on a stand. She moves a 4 cm tall candle along the principal axis and observes the image each time.
(a) Where must the candle be placed to get a real, inverted image of the same size? Justify with one line.
(b) The candle is now placed at 30 cm from the mirror. Using the mirror formula, find the position, magnification and size of the image.
(c) State the nature of the image when the candle is placed beyond the centre of curvature.
Marking Scheme (4M Case): Part(a) Understanding 1M · Part(b) Analysis 1.5M · Part(c) Application 1.5M
(a) Same size position
At the centre of curvature: u = 2f = 40 cm, where object and image are of the same size (both at C).
(b) Given data
f = −20 cm, u = −30 cm, h = 4 cm
(b) Mirror formula
1/v = 1/f − 1/u = (−1/20) − (−1/30) = −1/60 ⇒ v = −60 cm
(b) Magnification
m = −v/u = −(−60)/(−30) = −2 ⇒ h′ = 2 × 4 = 8 cm
(c) Beyond centre of curvature
The image lies between C and F — real, inverted and diminished.
✅ (b) Image 60 cm in front, m = −2, height 8 cm.
Case study. A convex lens of power +2.0 D is used to focus sunlight.
(a) Find the focal length of the lens.
(b) A 2 cm tall object is placed at 50 cm from the lens. Locate the image with the lens formula and find the magnification.
(c) State one real-life use of a convex lens.
— OR / अथवा
OR case. A concave lens of focal length 10 cm produces a virtual image 5 cm from the lens for a real object.
(a) State where the object is and its distance for this case.
(b) Find the magnification produced by a concave lens when |v| = 5 cm and |u| = 10 cm.
(c) Give two uses of a concave lens.
[Note: Attempt any ONE]
Marking Scheme (4M Case): Part(a) Understanding 1M · Part(b) Analysis 1.5M · Part(c) Application 1.5M
Option 1 — Convex lens, P = +2.0 D
(a) Focal length
f = 1/P = 1/2.0 = 0.5 m = 50 cm.
(b) Given data
f = +50 cm, u = −50 cm, h = 2 cm
(b) Lens formula
1/v − 1/u = 1/f ⇒ 1/v = 1/50 + (−1/50) = 0 ⇒ v at infinity
(b) Reading
At u = −50 cm = −f, the image is at infinity (rays emerge parallel).
(c) Use
Magnifying glass, camera lens or telescope objective.
Option 2 — Concave lens, f = −10 cm
(a) Object position
Lens formula 1/v − 1/u = 1/f with v = −5 cm gives 1/u = 1/v − 1/f = (−1/5) − (−1/10) = −1/10 ⇒ u = −10 cm, so the object is 10 cm in front.
(b) Magnification
m = v/u = (−5)/(−10) = +1/2 (erect, diminished).
(c) Uses
Spectacles for myopia, and shrinking lenses in peepholes.
✅ Option 1: f = 50 cm; object at focus gives image at infinity. ✅ Option 2: object at 10 cm, m = +1/2.
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