Optics: 20-Question Learning Review

How to use this review

Attempt this free, original learning review after the three optics lessons: Reflection and Mirrors; Refraction and Lenses; Human Eye, Dispersion and Optical Effects. There are 20 questions, each worth one mark, with exactly one correct option. The review is untimed, with no negative marking and no pass cutoff. A suggested 25 minutes is only a study aid, not an official examination duration. This is not a full RRB CBT simulation and does not reproduce official CBT scoring.

Use real upright objects unless a question says otherwise. Mirror and lens calculations use the paraxial model (rays close to the principal axis); lenses are thin and in air. Incident light travels from left to right. Under the Cartesian convention, distances to the right are positive and distances to the left negative, measured from the mirror pole or the lens optical centre. Thus u < 0 for the real objects used here. The sign of v must be interpreted for the particular mirror or lens, not guessed from the word “real”. Angles of incidence, reflection and refraction are measured from the normal.

Before viewing an explanation, record your choice and one reason or calculation. Then identify any error: normal versus surface, object position, sign convention, wrong mirror/lens formula, unit conversion, image nature, or confusion between refraction, dispersion and scattering. Revisit the relevant lesson and retry the missed question without looking at the answer.

Questions

1. A ray strikes a plane mirror at an angle of 28° to the mirror surface. What is the angle of reflection, measured from the normal?

A. 28° B. 56° C. 62° D. 90°

2. Parallel rays fall on a rough wall and are reflected in different directions. Which statement best explains this?

A. The local normals differ, but each reflected ray obeys the laws of reflection. B. The angle of reflection is unrelated to the angle of incidence on a rough surface. C. The wall sends each ray through its surface without reflection. D. Each reflected ray must remain parallel to every other reflected ray.

3. A person is initially 1.5 m in front of a fixed plane mirror. The person moves 0.4 m directly towards it. By how much does the distance between the person and their image decrease?

A. 0.4 m B. 0.8 m C. 1.1 m D. 2.2 m

4. A real object is placed between the principal focus F and the centre of curvature C of a concave mirror. Which description of its image is correct?

A. Virtual, erect and diminished, behind the mirror B. Real, inverted and diminished, between F and C C. Real, inverted and the same size, at C D. Real, inverted and enlarged, beyond C

5. A concave mirror has f = −16 cm. A real object is at u = −48 cm. Which pair gives the image distance v and signed magnification m?

A. v = +24 cm; m = +1/2 B. v = −24 cm; m = −1/2 C. v = −12 cm; m = −1/4 D. v = −24 cm; m = +1/2

6. A spherical mirror forms an image of a real upright object with signed magnification m = +2. Which statement is correct?

A. The image is real, inverted and twice as tall as the object. B. The image is virtual, erect and half as tall as the object. C. The image is virtual, erect and twice as tall as the object. D. The image is real, erect and the same height as the object.

7. A convex mirror has f = +24 cm and a real object at u = −48 cm. Where is the image formed?

A. 16 cm behind the mirror B. 16 cm in front of the mirror C. 48 cm behind the mirror D. 48 cm in front of the mirror

8. A concave spherical mirror has a radius of curvature of magnitude 36 cm. In the paraxial approximation, what is its signed focal length?

A. +18 cm B. −36 cm C. −72 cm D. −18 cm

9. A light ray enters ordinary glass from air at an angle of 40° to the normal. The glass has a higher refractive index than air. What happens to the transmitted ray?

A. It bends away from the normal and its speed decreases. B. It bends towards the normal and its speed increases. C. It bends towards the normal and its speed decreases. D. It remains undeviated and its speed stays the same.

10. Light enters glass from air exactly along the normal. Which statement correctly describes the transmitted light?

A. Its direction is unchanged, its speed decreases, and its frequency is unchanged. B. Its direction bends towards the normal, and its speed is unchanged. C. Its direction is unchanged, its speed decreases, and its frequency decreases. D. Its direction and speed are both unchanged because the incidence is normal.

11. A transparent medium has absolute refractive index n = 1.25. Taking the speed of light in vacuum as 3.0 × 10⁸ m/s, what is the speed of light in this medium?

A. 3.75 × 10⁸ m/s B. 2.4 × 10⁸ m/s C. 1.25 × 10⁸ m/s D. 3.0 × 10⁸ m/s

12. A ray passes obliquely through a rectangular glass slab with parallel faces, with air on both sides. Which description of the emergent ray is correct?

A. It is always perpendicular to the exit face. B. It is not parallel to the incident ray because the two faces act like a prism. C. It exactly retraces the incident ray towards the source. D. It is parallel to the incident ray but laterally displaced.

13. A thin convex lens in air has f = +24 cm and a real object at u = −72 cm. Which pair gives its image distance v and signed magnification m?

A. v = +36 cm; m = −1/2 B. v = −36 cm; m = +1/2 C. v = +18 cm; m = −1/4 D. v = +36 cm; m = +1/2

14. A thin concave lens in air has f = −30 cm and a real object at u = −60 cm. Which description of the image is correct?

A. v = +20 cm; real and inverted B. v = −60 cm; virtual and the same size C. v = −20 cm; virtual, erect and one-third as tall D. v = −20 cm; virtual, inverted and one-third as tall

15. A thin concave lens in air has a signed focal length of −125 cm. What is its power?

A. −0.008 D B. +0.80 D C. −125 D D. −0.80 D

16. A normal eye shifts focus from a nearby object to a distant object, both within its clear-vision range. Which change helps focus the distant object on the retina?

A. The ciliary muscles contract, the lens becomes thicker and its focal length decreases. B. The ciliary muscles relax, the lens becomes thinner and its focal length increases. C. The retina moves backwards to meet the image while the lens remains unchanged. D. The pupil itself forms the image by replacing the eye lens.

17. In the basic school-level model, which pair correctly matches suitable corrective lenses for myopia and hypermetropia, respectively?

A. Convex; concave B. Convex; convex C. Concave; concave D. Concave; convex

18. A first glass prism separates white light into a spectrum. A second identical prism is suitably aligned in the reversed orientation so that all these colours pass through it and overlap again. What can emerge from the second prism?

A. White light, because the component colours have been recombined B. Only violet light, because it was deviated most by the first prism C. No light, because one prism permanently destroys the colours D. Only red light, because the second prism must absorb every other colour

19. What best explains why a distant star appears to twinkle when viewed through Earth’s atmosphere?

A. The star repeatedly switches its light production on and off over very short intervals. B. A fixed plane mirror in the atmosphere reflects alternate rays away. C. Changing atmospheric conditions vary the refraction and the amount of starlight reaching the eye. D. The eye lens splits white starlight into colours by the same process as a glass prism.

20. On a clear day, why does the sky away from the Sun usually appear blue?

A. Air molecules scatter red light more strongly than blue light. B. Air molecules scatter shorter-wavelength blue light more strongly than longer-wavelength red light. C. A single atmospheric prism separates sunlight into one fixed blue band across the whole sky. D. The blue colour of the oceans is the main cause of the sky’s colour everywhere.

Explained answers

1. C — The normal is perpendicular to the mirror, so the angle of incidence is 90° − 28° = 62°. The reflection law gives r = i = 62°. A measures from the surface instead of the normal. B doubles the given angle without finding i. D gives the surface–normal angle, not the reflection angle.

2. A — Roughness changes the direction of the normal from one point to another. At each point, i = r relative to that local normal. The different normals produce diffuse reflection. B incorrectly discards the reflection law. C describes transmission, not the stated observation. D would fit ideal reflection of parallel rays from a smooth plane surface, not this rough wall.

3. B — The image is as far behind a plane mirror as the object is in front. The initial separation is 2 × 1.5 = 3.0 m. The new object distance is 1.5 − 0.4 = 1.1 m, so the new separation is 2.2 m. The decrease is 3.0 − 2.2 = 0.8 m. A counts only the person’s motion; C is the new person–mirror distance; D is the final person–image separation, not its decrease.

4. D — For a concave mirror, an object between F and C forms an image beyond C that is real, inverted and enlarged. A is the usual convex-mirror case for a real object. B belongs to a concave mirror with the object beyond C. C belongs to an object at C. The object position is therefore essential; the mirror name alone is insufficient.

5. B — For a mirror, 1/v = 1/f − 1/u = −1/16 + 1/48 = −1/24 cm⁻¹, so v = −24 cm. Then m = −v/u = −(−24)/(−48) = −1/2: a real, inverted image of half the object’s height. A reverses both signs. C results from incorrectly adding 1/u to 1/f. D gets v right but uses the lens magnification formula v/u instead of the mirror formula −v/u.

6. C — The positive sign of m means the image is erect; |m| = 2 means twice the object’s height. With a real object u < 0 and mirror m = −v/u, m > 0 gives v > 0, behind the mirror, so the image is virtual. A ignores the sign. B confuses 2 with its reciprocal. D gets both the image nature and size wrong. A concave mirror with the object inside its focal distance can produce this image.

7. A — 1/v = 1/f − 1/u = 1/24 + 1/48 = 1/16 cm⁻¹, giving v = +16 cm, behind the mirror. Also m = −16/(−48) = +1/3, consistent with an erect, diminished virtual image between the pole and focus. B assigns the wrong side to positive v. C comes from subtracting the magnitudes 1/24 − 1/48 instead of using signed u. D combines that magnitude error with the wrong side.

8. D — For paraxial rays, f = R/2. The centre of curvature of this concave mirror is on the incident-light side, so R = −36 cm and f = −18 cm. A misses the concave-mirror sign. B sets focal length equal to the radius. C doubles the radius instead of halving it. The radius is measured from the pole to the centre of curvature.

9. C — A higher refractive index means a lower speed of light. For this oblique entry into the higher-index medium, the refracted angle is smaller than the incident angle, so the ray bends towards the normal. A gets the bending direction wrong. B gets the speed change wrong. D incorrectly treats oblique incidence as normal incidence and also misses the speed change.

10. A — At normal incidence, i = r = 0°, so there is no change of direction. Light still travels more slowly in glass, and its frequency remains fixed at the boundary. Its wavelength decreases because speed = frequency × wavelength. B invents a bend and misses the speed change. C incorrectly changes frequency. D confuses no bending with no change in speed.

11. B — n = c/speed, so speed = c/n = (3.0 × 10⁸)/1.25 = 2.4 × 10⁸ m/s. A multiplies by n instead of dividing. C treats the numerical value of the index as a speed coefficient. D ignores the index and leaves the vacuum speed unchanged. The result being less than c is a useful check.

12. D — The ray bends towards the normal on entering glass and away from the normal on leaving. Because the faces are parallel and the outside medium is the same, the emergent direction is parallel to the incident direction; oblique passage produces a lateral displacement. A wrongly imposes normal emergence. B overlooks the parallel faces, unlike a prism’s inclined faces. C describes returning towards the source, not transmission through the slab.

13. A — For a lens, 1/v = 1/f + 1/u = 1/24 − 1/72 = 1/36 cm⁻¹. Thus v = +36 cm and m = v/u = 36/(−72) = −1/2. The image is real, inverted and half the object’s height, on the opposite side of the lens. B reverses the signs. C uses the mirror equation instead of the lens equation. D uses the mirror magnification formula −v/u instead of the lens formula v/u.

14. C — 1/v = 1/f + 1/u = −1/30 − 1/60 = −1/20 cm⁻¹, so v = −20 cm. The image lies on the object’s side and is virtual. Lens magnification m = v/u = (−20)/(−60) = +1/3 makes it erect and diminished. A assigns the wrong sign and image nature. B comes from using the mirror equation. D gets the position right but misreads positive magnification as inversion.

15. D — First convert f = −125 cm = −1.25 m. Power P = 1/f = 1/(−1.25) = −0.80 D. The negative power agrees with a diverging concave lens in air. A takes the reciprocal of centimetres without converting to metres. B drops the negative sign. C repeats the focal-length number instead of taking its reciprocal in metres.

16. B — For a distant object, the ciliary muscles relax and the eye lens becomes less curved and thinner. Its focal length increases and its converging power decreases, allowing the image to remain on the retina. A describes focusing on a nearby object. C confuses accommodation with moving the retina. D confuses the pupil’s role in admitting light with the lens system’s focusing role.

17. D — In myopia, rays from a distant object would focus in front of the retina; a suitable concave lens reduces convergence before light enters the eye. In hypermetropia, rays from a near object would focus behind the retina; a suitable convex lens provides extra convergence. A swaps the two corrections. B wrongly adds convergence for myopia. C wrongly reduces convergence for hypermetropia. This is a model question, not a prescription for an individual.

18. A — The first prism separates colours already present in white light by refracting them through different angles. A suitably aligned reversed prism can recombine that spectrum into white light. B confuses greater deviation with survival of only one colour. C mistakes separation for destruction. D invents selective absorption; an ordinary prism does not have to remove every colour except red. The alignment condition matters: simply placing any second prism nearby does not guarantee recombination.

19. C — A distant star is approximately a point source. Changing refractive-index conditions in the atmosphere make its ray paths and apparent brightness fluctuate, producing twinkling. A assigns the ordinary atmospheric effect to rapid changes in the star itself. B invents a reflecting surface. D confuses twinkling with prism dispersion. The explanation concerns atmospheric refraction, not a failure of the reflection law.

20. B — Air molecules and very fine particles scatter shorter visible wavelengths more strongly than longer ones. Scattered light from the blue end reaches our eyes from directions away from the Sun, making the clear sky appear blue. A reverses the wavelength preference. C confuses scattering in many directions with prism dispersion into a spectrum. D is not the explanation: a blue sky is also seen far from oceans.

Source note

Scientific checks: NIOS Secondary Science and Technology, Light Energy (OPT-S01); NCERT Class X, Light – Reflection and Refraction (OPT-S04); NCERT Class X, The Human Eye and the Colourful World (OPT-S03). NCERT Class XII Ray Optics (OPT-S02) is used only to cross-check the paraxial mirror model and signs, not to raise the review level. The question wording, numerical choices and explanations are original; these are not previous-year questions and no topic-frequency prediction is made.

Quick reference

Review checklist

• Measure ray angles from the normal, not the surface
• Mirrors: 1/f = 1/v + 1/u; signed magnification m = −v/u
• Thin lenses in air: 1/f = 1/v − 1/u; signed magnification m = v/u
• Absolute refractive index n = c/(speed in medium); c is the vacuum speed
• Lens power P = 1/f only when f is in metres and P is in dioptres
• Explain the image or optical effect as well as selecting an answer