Previous year questions: Ray Optics And Optical Instruments
Ray Optics and Optical Instruments28 min readLesson 6 of 6
Q1 (CBSE Board 2022, 3 marks)
A glass converging lens (refractive index 1.5) has faces with radii of curvature 10 cm and 20 cm. Work out its focal length first in air and then inside a liquid of refractive index 1.25.
Answer: With the first face of radius 10 cm positive and the second face of radius 20 cm negative in the sign convention, the lens maker formula in air gives one over f equal to 0.5 times (1/10 plus 1/20), which is 0.5 times 0.15, so f equals 40/3, about 13.3 cm. In the liquid the relative refractive index is 1.5/1.25, equal to 1.2, so one over f equals 0.2 times 0.15, equal to 0.03, giving f equal to 100/3, about 33.3 cm. The lens remains converging in both cases because the relative index stays above one.
Q2 (CBSE Board 2022, 3 marks)
A point bulb sits at the bottom of a water tank 70 cm deep. Taking water's refractive index as 4/3, work out the area of the surface patch through which its light can escape into air.
Answer: Light emerges only within the circle where the angle at the surface is less than the critical angle. For a point source at depth d the radius of that circle is d divided by the square root of (n squared minus 1). With n equal to 4/3 the square root of (16/9 minus 1) is the square root of 7/9, so the radius is 70 times 3 divided by root 7, about 79.4 cm. The area is pi times the square of the radius, about 1.98 square metres. The circle is the region where the angle of incidence at the surface stays below the critical angle. The bright patch is a circle on the surface whose rim is set by the critical angle.
Q3 (CBSE Board 2022, 3 marks)
A small telescope uses an objective of focal length 150 cm with an eyepiece of focal length 6 cm. For the final image at infinity, work out the angular magnification and the separation between the lenses.
Answer: For a final image at infinity the magnifying power is the objective focal length divided by the eyepiece focal length, which is 150/6, equal to 25. The tube length is the sum of the two focal lengths, 150 plus 6, equal to 156 cm. The negative sign of the magnification is not needed here because the question asks for the magnitude. The tube length follows from the way the foci of the two lenses coincide in normal adjustment. In normal adjustment the two focal points coincide, so the tube length is the sum of the focal lengths.
Q4 (CBSE Board 2023, 2 marks)
A symmetrical double-convex lens has power 4 D and is cut from glass of refractive index 1.5. If both surfaces have equal curvature, find the radius of curvature of each surface.
Answer: Power 4 D means the focal length is 0.25 m, or 25 cm. For an equiconvex lens the lens maker formula gives one over f equal to (n minus 1) times 2 over R. Substituting one over 25 equals 0.5 times 2 over R gives R equal to 25 cm. Both faces have a radius of curvature of 25 cm. The formula gives the radius directly once the focal length is converted to metres and back. The equal radii make the two surface terms identical, which is why the factor of two appears.
Q5 (CBSE Board 2023, 2 marks)
Give two reasons why an astronomical telescope is often built with a mirror objective rather than a lens objective.
Answer: A reflecting telescope uses a concave mirror objective, so it has no chromatic aberration, unlike a refracting telescope whose lens disperses different colours by different amounts. A mirror can also be built much larger than a lens without the problems of sagging and supporting a heavy glass disc, so it gives a larger aperture, brighter images and better resolution. Mentioning chromatic aberration and the size of the aperture covers the two expected reasons. A mirror avoids colour fringing, and it can be made larger than a lens without sagging.
Q6 (CBSE Board 2023, 1 mark)
A convex lens of focal length f forms a real image of a real object. Which is the smallest possible separation between the object and the image: f, 2f, 3f or 4f?
Answer: For a convex lens the distance between the object and its real image is smallest when the object and the image are at equal distances on the two sides, that is at twice the focal length from the lens. The minimum separation is then 4f. The correct choice is 4f. The equality case is symmetric, which is why the separation comes out as twice the object distance. The symmetric arrangement places the object at twice the focal length, giving a separation of four times f.
Q7 (CBSE Board 2025, 2 marks)
An equiconvex lens is made of glass of refractive index 1.55. If its focal length is 15.0 cm, calculate the radius of curvature of each face.
Answer: For an equiconvex lens the lens maker formula becomes one over f equals (n minus 1) times 2 over R. Substituting one over 15 equals 0.55 times 2 over R gives R equal to 0.55 times 30, which is 16.5 cm. Both surfaces have the same radius. The relative error in the radius is what examiners check here, so keep the multiplication exact. The radius follows directly from the focal length and the relative index.
Q8 (CBSE Board 2025, 2 marks)
A plano-convex lens is made of glass of refractive index 1.5 with a curved face of radius 10 cm. Find its focal length after the lens is placed in a liquid of refractive index 1.25.
Answer: For a plano-convex lens the lens maker formula with the plane surface contribution zero gives one over f equal to (n relative minus 1) divided by R. The relative refractive index is 1.5/1.25, equal to 1.2. Then one over f equals 0.2/10, which is 0.02, so f equals 50 cm. The positive value shows the lens still converges in the liquid. The relative refractive index is the only quantity that changes when the surrounding medium changes. The plano-convex form keeps only one surface term in the formula.
Q9 (CBSE SQP 2021-22, 3 marks)
A glass convex lens of refractive index 1.5 has focal length 20 cm in air. Work out the focal length after the lens is put into a medium of refractive index 1.25, and say whether it still converges.
Answer: The lens maker formula shows that the focal length is proportional to (n minus 1) for the same radii, with n measured relative to the surrounding medium. In air the factor is 0.5 and in the liquid it is 1.5/1.25 minus 1, equal to 0.2. The new focal length is therefore 20 times 0.5 divided by 0.2, which is 50 cm. It is positive, so the lens remains converging. The ratio of the two lens factors gives the new focal length without recomputing the radii. The ratio of the two factors gives the new focal length without recalculating the radii. The positive sign is part of what the question asks, so state it clearly.
Q1 (CBSE Board 2022, 3 marks)
A glass converging lens (refractive index 1.5) has faces with radii of curvature 10 cm and 20 cm. Work out its focal length first in air and then inside a liquid of refractive index 1.25.
Answer: With the first face of radius 10 cm positive and the second face of radius 20 cm negative in the sign convention, the lens maker formula in air gives one over f equal to 0.5 times (1/10 plus 1/20), which is 0.5 times 0.15, so f equals 40/3, about 13.3 cm. In the liquid the relative refractive index is 1.5/1.25, equal to 1.2, so one over f equals 0.2 times 0.15, equal to 0.03, giving f equal to 100/3, about 33.3 cm. The lens remains converging in both cases because the relative index stays above one.