About This Section

This section collects CBSE and other board style questions on Light – Reflection and Refraction, grouped by marks. Practise the numericals with full sign conventions and the ray diagrams by hand.

Study Strategy:

  1. Write given data with signs; state the formula before substituting.
  2. 1-mark = direct fact/definition; 3-mark = numerical or diagram; 5-mark = derivation-style numerical + ray diagram.
  3. Always state the nature of the image.
  4. Solve all NCERT exercises.

1-Mark Questions (MCQ / Very Short)

Q1. [CBSE] The mirror formula is — (a) 1/v - 1/u = 1/f (b) 1/v + 1/u = 1/f (c) v + u = f (d) 1/v + 1/u = 2/f Solution: (b) 1/v + 1/u = 1/f.

Q2. [CBSE] The SI unit of power of a lens is the — (a) metre (b) dioptre (c) watt (d) newton Solution: (b) dioptre.

Q3. [CBSE] A convex mirror always forms an image that is — (a) real, inverted (b) virtual, erect, diminished (c) real, enlarged (d) virtual, enlarged Solution: (b) virtual, erect, diminished.

Q4. [CBSE] Light bends towards the normal when it goes from — (a) denser to rarer (b) rarer to denser (c) same media (d) glass to air Solution: (b) rarer to denser.

1-Mark Questions (continued)

Q5. [CBSE] State the relation between radius of curvature and focal length of a spherical mirror. Solution: R=2fR = 2f.

Q6. [CBSE] Which mirror is used as a vehicle headlight reflector? Solution: A concave mirror (source at focus gives a parallel beam).

Q7. [CBSE] Define the power of a lens. Solution: Power is the reciprocal of the focal length (in metres): P=1/fP = 1/f; SI unit dioptre.

Q8. [CBSE] The magnification of a plane mirror is +1. What does it indicate? Solution: The image is virtual, erect and of the same size as the object.

2-Mark Questions

Q9. [CBSE] State the two laws of reflection of light. Solution: (i) The angle of incidence equals the angle of reflection. (ii) The incident ray, the normal at the point of incidence and the reflected ray lie in the same plane.

Q10. [CBSE] The refractive index of glass is 1.5. Find the speed of light in glass (c=3×108c = 3\times10^8 m/s). Solution: v=c/n=(3×108)/1.5=2×108v = c/n = (3\times10^8)/1.5 = 2\times10^8 m/s.

2-Mark Questions (continued)

Q11. [CBSE] Why does a ray emerging from a rectangular glass slab run parallel to the incident ray? Solution: The bending towards the normal at the first face and away from the normal at the second (parallel) face are equal and opposite, so the emergent ray is parallel to the incident ray (only laterally shifted).

Q12. [CBSE] A concave lens has power -2 D. Find its focal length and type. Solution: f=1/P=1/(−2)=−0.5f = 1/P = 1/(-2) = -0.5 m; it is a concave (diverging) lens.

Q13. [CBSE] State two uses of a concave mirror. Solution: As a shaving/dentist's mirror (enlarged erect image) and in vehicle headlights/search-lights (parallel beam)/solar furnaces.

3-Mark Questions

Q14. [CBSE] An object 4.0 cm tall is placed 25 cm from a concave mirror of focal length 15 cm. Find the image distance, size and nature. Solution: u=−25u=-25, f=−15f=-15. 1v=−115+125=−275\dfrac{1}{v}=-\dfrac{1}{15}+\dfrac{1}{25}=\dfrac{-2}{75}, so v=−37.5v=-37.5 cm (real). h′=−vuh=−6.0h' = -\dfrac{v}{u}h = -6.0 cm. Real, inverted, enlarged, 6.0 cm tall.

Q15. [CBSE] State the New Cartesian Sign Convention (any three rules) for spherical mirrors. Solution: (i) The object is placed to the left of the mirror. (ii) All distances are measured from the pole. (iii) Distances to the right of the pole are positive, to the left negative; heights above the axis positive, below negative.

3-Mark Questions (continued)

Q16. [CBSE] Draw a ray diagram and describe the image when an object is placed between F and C of a concave mirror. Solution: Draw two rays (one parallel to the axis reflecting through F, one through C returning on itself). The image forms beyond C, and is real, inverted and enlarged.

Q17. [CBSE] A convex lens forms a real, inverted image at 30 cm when the object is at 15 cm. Find the focal length and magnification. Solution: u=−15u=-15, v=+30v=+30. 1f=1v−1u=130+115=330\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{30}+\dfrac{1}{15}=\dfrac{3}{30}, so f=+10f=+10 cm. m=vu=30−15=−2m=\dfrac{v}{u}=\dfrac{30}{-15}=-2 (real, inverted, 2× enlarged).

Q18. [CBSE] Distinguish between a real and a virtual image (two points each). Solution: A real image is formed by actually converging rays, can be caught on a screen, and is inverted. A virtual image is formed by apparently diverging rays, cannot be caught on a screen, and is erect.

5-Mark Questions

Q19. [CBSE] (a) Draw ray diagrams for image formation by a concave mirror when the object is (i) at C and (ii) between P and F. (b) State the nature of each image. Solution: (a) Draw the two cases using standard rays. (b) (i) Object at C → image at C: real, inverted, same size. (ii) Object between P and F → image behind the mirror: virtual, erect, enlarged.

Q20. [CBSE] An object 2.0 cm tall is placed 15 cm from a convex lens of focal length 10 cm. Find the position, size and nature of the image, with a ray diagram. Solution: u=−15u=-15, f=+10f=+10. 1v=110−115=130\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{1}{30}, so v=+30v=+30 cm. m=vu=−2m=\dfrac{v}{u}=-2, h′=−4.0h'=-4.0 cm. Real, inverted, enlarged (2×), 4 cm tall, 30 cm on the other side. Draw two construction rays to confirm.

5-Mark Questions (continued)

Q21. [CBSE] Define refractive index. The refractive index of water is 1.33 and of glass 1.52. (a) In which does light travel faster? (b) Find the speed of light in water (c=3×108c=3\times10^8 m/s). Solution: Refractive index n=c/vn = c/v = (speed in vacuum)/(speed in medium). (a) Light is faster in water (lower refractive index). (b) v=c/n=(3×108)/1.33=2.26×108v = c/n = (3\times10^8)/1.33 = 2.26\times10^8 m/s.

Q22. [CBSE] (a) State the laws of refraction. (b) A convex lens of focal length 10 cm is used as a magnifying glass with an object 6 cm from it. Find the image distance and magnification. Solution: (a) (i) Incident ray, refracted ray and normal lie in one plane. (ii) Snell's law: sin⁡isin⁡r\dfrac{\sin i}{\sin r} = constant (refractive index). (b) u=−6u=-6, f=+10f=+10. 1v=110−16=3−530=−230\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{6}=\dfrac{3-5}{30}=\dfrac{-2}{30}, so v=−15v=-15 cm. m=vu=−15−6=+2.5m=\dfrac{v}{u}=\dfrac{-15}{-6}=+2.5. Image virtual, erect, enlarged (a magnifying glass).