IGCSE Physics 3.2 Light MCQs – 30 Practice Questions with Answers (Cambridge 0625)
Free practice MCQs for Cambridge IGCSE Physics (0625) Section 3.2, Light, covering reflection, refraction, total internal reflection, lenses and dispersion for both Core and Supplement material. Work through each question, then open the Show Answer toggle to check the fully worked explanation.
Formula recap: angle of incidence = angle of reflection | n = sin i ÷ sin r | sin C = 1 ÷ n | all angles are measured from the normal. Need the full revision sheet? See the IGCSE Physics Formula Sheet.
Reflection & Plane Mirrors
Q1. A ray of light strikes a plane mirror. The angle between the incident ray and the normal is 40°. What is the angle of reflection?
- (A) 20°
- (B) 40°
- (C) 50°
- (D) 80°
Show Answer
Answer: B – The law of reflection states that the angle of reflection equals the angle of incidence. Both angles are measured from the normal, so the angle of reflection is also 40°.
Q2. A ray of light makes an angle of 30° with the surface of a plane mirror. What is the angle of reflection?
- (A) 30°
- (B) 45°
- (C) 60°
- (D) 90°
Show Answer
Answer: C – Angles are measured from the normal, not from the surface. The angle of incidence is 90° − 30° = 60°, so the angle of reflection is also 60°. Using the angle with the surface directly is the classic trap here.
Q3. The angle of incidence of a ray striking a plane mirror is 55°. What is the angle between the incident ray and the reflected ray?
- (A) 110°
- (B) 55°
- (C) 35°
- (D) 125°
Show Answer
Answer: A – The angle of reflection equals the angle of incidence, 55°. The two rays lie on opposite sides of the normal, so the angle between them is 55° + 55° = 110°.
Q4. Which statement about the image formed by a plane mirror is correct?
- (A) It is real and can be formed on a screen
- (B) It is larger than the object
- (C) It is turned upside down
- (D) It is virtual, upright and laterally inverted
Show Answer
Answer: D – A plane mirror image is virtual (the rays only appear to meet behind the mirror, so it cannot be caught on a screen), upright, the same size as the object, and laterally inverted – left and right are swapped.
Q5. An object is placed 2.5 m in front of a plane mirror. How far is the image from the object?
- (A) 1.25 m
- (B) 2.5 m
- (C) 5.0 m
- (D) 7.5 m
Show Answer
Answer: C – In a plane mirror the image is as far behind the mirror as the object is in front of it, so the image is 2.5 m behind the mirror. The distance between object and image is 2.5 m + 2.5 m = 5.0 m.
Q6. The writing on the front of an ambulance is printed in mirror writing (reversed left-to-right). Why?
- (A) So that the writing looks bigger from a distance
- (B) So that drivers in front read it correctly in their rear-view mirrors
- (C) So that the writing can be seen at night
- (D) So that the writing is the right way up for the driver of the ambulance
Show Answer
Answer: B – A mirror image is laterally inverted. The reversed writing is swapped left-to-right again when seen in the rear-view mirror of a car in front, so it reads correctly and the driver can move aside.
Q7. A polished metal sheet gives a clear reflected image, but a painted wall does not give a clear image even though it reflects light. This is because the wall surface…
- (A) reflects the rays irregularly in many different directions
- (B) absorbs all the light that falls on it
- (C) does not obey the law of reflection
- (D) refracts the light into the wall
Show Answer
Answer: A – The wall is rough, so the normal points in different directions at different places. Each ray still obeys the law of reflection, but the reflected rays scatter in many directions – this is diffuse reflection, so no clear image forms.
Q8. In a simple periscope, how are the two plane mirrors arranged?
- (A) Standing vertically, facing each other
- (B) Both lying horizontally, one above the other
- (C) At right angles to each other, forming a V shape
- (D) Parallel to each other, each at 45° to the line of sight
Show Answer
Answer: D – The top mirror, tilted at 45°, reflects light from the object straight down the tube. The bottom mirror, parallel to the first and also at 45°, reflects it horizontally into the eye, giving an upright image.
Q9. A student walks straight towards a plane mirror at 1.0 m/s. At what speed does her image move towards her?
- (A) 0.5 m/s
- (B) 2.0 m/s
- (C) 1.0 m/s
- (D) 4.0 m/s
Show Answer
Answer: B – The image is always the same distance behind the mirror as the student is in front, so the image also approaches the mirror at 1.0 m/s. The gap between student and image closes at 1.0 + 1.0 = 2.0 m/s.
Q10. What happens to the image in a plane mirror when the object is moved further away from the mirror?
- (A) The image becomes larger
- (B) The image becomes smaller
- (C) The image stays the same size but is further behind the mirror
- (D) The image becomes real
Show Answer
Answer: C – A plane mirror always produces an image the same size as the object, however far away the object is. Moving the object back simply moves the virtual image to the same new distance behind the mirror.
Refraction & Total Internal Reflection
Q11. A ray of light passes from air into glass and bends towards the normal. This happens because…
- (A) light slows down as it enters the glass
- (B) light speeds up as it enters the glass
- (C) the frequency of the light increases in the glass
- (D) the wavelength of the light increases in the glass
Show Answer
Answer: A – Light travels more slowly in glass than in air. The change of speed at the boundary makes the ray change direction – it bends towards the normal when it slows down entering the denser medium.
Q12. A ray of light passes from air into glass. The angle of incidence is 45° and the angle of refraction is 28°. Use sin 45° = 0.707 and sin 28° = 0.469. What is the refractive index of the glass? [Supplement]
- (A) 0.47
- (B) 1.5
- (C) 0.66
- (D) 2.1
Show Answer
Answer: B – n = sin i ÷ sin r = 0.707 ÷ 0.469 = 1.51, which is approximately 1.5. Option C is the inverted ratio, a common slip.
Q13. The speed of light in a vacuum is 3.0 × 108 m/s. In a certain material the speed of light is 2.0 × 108 m/s. What is the refractive index of the material? [Supplement]
- (A) 0.67
- (B) 1.2
- (C) 2.0
- (D) 1.5
Show Answer
Answer: D – Refractive index can also be found from speeds: n = speed in vacuum ÷ speed in the material = 3.0 × 108 ÷ 2.0 × 108 = 1.5. The greater the refractive index, the more the light is slowed.
Q14. A swimming pool looks shallower than it really is. This is because light travelling from the bottom of the pool to the eye…
- (A) bends towards the normal as it passes from water into air
- (B) is totally internally reflected at the surface
- (C) bends away from the normal as it passes from water into air
- (D) travels more slowly in air than in water
Show Answer
Answer: C – Light speeds up as it leaves the water, so it bends away from the normal. The eye traces the rays back in straight lines, so the bottom of the pool appears raised – the pool looks shallower than its real depth.
Q15. Total internal reflection can take place when a ray of light…
- (A) travels in a denser medium towards a less dense medium, with an angle of incidence greater than the critical angle
- (B) travels from air into glass at a large angle of incidence
- (C) travels in a denser medium at any angle of incidence
- (D) travels from a less dense medium towards a denser medium
Show Answer
Answer: A – Both conditions must hold: the light must be travelling from the optically denser medium towards the less dense one, AND its angle of incidence must exceed the critical angle. It then reflects back inside the denser medium, with nothing refracting out.
Q16. The critical angle for glass is about 42°. A ray of light inside the glass strikes the glass-air boundary at 50° to the normal. What happens to the ray?
- (A) It refracts out of the glass, bending towards the normal
- (B) It refracts out of the glass, bending away from the normal
- (C) It is absorbed by the surface of the glass
- (D) It is totally internally reflected back into the glass
Show Answer
Answer: D – The angle of incidence (50°) is greater than the critical angle (42°), and the ray is travelling from glass towards air. Both conditions for total internal reflection are met, so the whole ray reflects inside the glass.
Q17. Optical fibres are used in endoscopes and in communications. Light sent along a fibre reaches the far end even when the fibre is bent because…
- (A) the light speeds up at each bend
- (B) the light is totally internally reflected many times inside the fibre
- (C) the glass of the fibre absorbs and re-emits the light
- (D) the light is refracted towards the normal at each surface
Show Answer
Answer: B – Inside the fibre the light always strikes the boundary at an angle greater than the critical angle, so it is totally internally reflected again and again. It zigzags along the fibre and emerges at the far end – which is how an endoscope lets a doctor see inside the body.
Q18. A student traces a ray through a rectangular glass block using a ray box. Which step is essential for measuring the angles of incidence and refraction correctly?
- (A) Measure both angles from the surface of the block
- (B) Place the block so the ray enters along the surface
- (C) Draw the normal at the entry point and measure both angles from the normal
- (D) Make the ray hit the exact centre of the block
Show Answer
Answer: C – The laws of refraction are stated in terms of angles from the normal (the line at 90° to the surface). Measuring from the surface instead is the most common source of wrong results in this experiment.
Q19. When light passes from air into glass, which quantity does NOT change?
- (A) The frequency of the light
- (B) The speed of the light
- (C) The wavelength of the light
- (D) The direction of travel (for a ray not along the normal)
Show Answer
Answer: A – The frequency is fixed by the source and stays the same in every medium. Because the speed decreases in glass and v = f × λ, the wavelength must decrease in the same proportion.
Q20. Diamond has a refractive index of 2.5. Using sin C = 1 ÷ n, what is the critical angle for diamond? (sin 24° = 0.40) [Supplement]
- (A) 42°
- (B) 24°
- (C) 30°
- (D) 66°
Show Answer
Answer: B – sin C = 1 ÷ 2.5 = 0.40, so C = 24°. This unusually small critical angle traps light inside a cut diamond, bouncing it around until it exits through the top – which is why diamonds sparkle.
Lenses & Dispersion
Q21. Which description matches a converging (convex) lens?
- (A) It is thinner at the centre than at the edges and spreads parallel rays apart
- (B) It is thicker at the edges than at the centre and brings parallel rays together
- (C) It is thicker at the centre than at the edges and brings parallel rays to a principal focus
- (D) It has the same thickness everywhere and leaves parallel rays parallel
Show Answer
Answer: C – A converging lens is fatter in the middle. Refraction at its two surfaces bends rays parallel to the principal axis so that they meet at the principal focus on the far side.
Q22. The principal focus of a converging lens is the point at which…
- (A) rays parallel to the principal axis meet after passing through the lens
- (B) every image is formed, wherever the object is placed
- (C) the centre of the lens is located
- (D) an object must be placed to give the largest possible image
Show Answer
Answer: A – Rays arriving parallel to the principal axis are refracted so that they all pass through the principal focus. The distance from the centre of the lens to this point is the focal length.
Q23. A single diverging lens is used on its own to view an object. The image formed is always…
- (A) real, inverted and magnified
- (B) virtual, upright and magnified
- (C) real, upright and diminished
- (D) virtual, upright and diminished
Show Answer
Answer: D – A diverging lens spreads parallel rays out as if they came from a virtual focus on the same side as the object. Whatever the object position, the image is virtual, upright and smaller than the object – which is why diverging lenses are used in spectacles for short sight.
Q24. An object is placed beyond 2F from a converging lens, as in a camera. The image formed is…
- (A) virtual, upright and magnified
- (B) real, inverted and diminished
- (C) real, inverted and the same size as the object
- (D) virtual, inverted and diminished
Show Answer
Answer: B – With the object beyond twice the focal length, the image forms between F and 2F on the other side of the lens. It is real (it can fall on the film or sensor of a camera), inverted and diminished.
Q25. A converging lens is used as a magnifying glass. To get a magnified, upright image, the object must be placed…
- (A) between the lens and its principal focus
- (B) exactly at the principal focus
- (C) exactly at 2F
- (D) beyond 2F
Show Answer
Answer: A – With the object closer to the lens than the focal length, the refracted rays diverge on the far side and only appear to meet on the object side. The eye sees a virtual image that is upright and magnified.
Q26. An object is placed exactly at 2F from a converging lens. The image formed is…
- (A) virtual, upright and the same size as the object
- (B) real, inverted and magnified
- (C) real, upright and diminished
- (D) real, inverted and the same size as the object, at 2F on the other side
Show Answer
Answer: D – Object at 2F is the in-between case: the image also forms at 2F on the opposite side, real, inverted and exactly the same size. Moving the object closer than 2F makes the image magnified; further away makes it diminished.
Q27. In the human eye, the lens forms an image on the retina. The image on the retina is…
- (A) real, upright and magnified
- (B) virtual, upright and diminished
- (C) real, inverted and magnified
- (D) real, inverted and diminished
Show Answer
Answer: D – The eye works like a camera with a fixed image distance (the depth of the eyeball). Most objects are far beyond 2F of the eye lens, so the image on the retina is real, inverted and diminished; the brain turns it the right way up.
Q28. A student finds the focal length of a converging lens by focusing a distant window onto a screen and measuring the lens-to-screen distance. This works because…
- (A) the image of the window is the same size as the window
- (B) rays from a very distant object arrive almost parallel, so they are focused at the principal focus
- (C) the window is exactly at 2F from the lens
- (D) a white screen always shows the image at the focal point
Show Answer
Answer: B – Light from a distant object arrives as an effectively parallel beam. A converging lens brings a parallel beam to its principal focus, so the sharp image on the screen is one focal length from the lens.
Q29. White light passes through a glass prism and is dispersed into a spectrum. Which colour is deviated (bent) the most by the prism?
- (A) Red
- (B) Green
- (C) Violet
- (D) Yellow
Show Answer
Answer: C – The refractive index of glass is slightly greater for violet light than for red light, so violet is slowed more and bent (deviated) the most. Red is deviated the least – that is why red is on the outside edge of a rainbow-coloured spectrum from a prism.
Q30. Which list gives the colours of the visible spectrum in order, from longest wavelength to shortest wavelength?
- (A) Red, orange, yellow, green, blue, indigo, violet
- (B) Violet, indigo, blue, green, yellow, orange, red
- (C) Red, yellow, green, orange, violet, blue, indigo
- (D) Blue, green, yellow, orange, red, indigo, violet
Show Answer
Answer: A – The spectrum runs red (longest wavelength, about 700 nm) down to violet (shortest wavelength, about 400 nm). This ordering also matches dispersion: red is deviated least and violet most by a prism.
Keep practising: Explore more free MCQs for every board on our Physics MCQs hub, and keep the IGCSE Physics Formula Sheet open while you revise – n = sin i ÷ sin r and the lens rules appear again and again in Paper 2.
All questions on this page are original practice questions written for QuaidianPoint. They are not Cambridge past-paper questions.