A symmetric biconvex lens made of glass (refractive index μ L = 1.5 ) has a focal length of f L = 20 cm in air. It is placed on a horizontal plane mirror. An object is placed on the principal axis

Question

A symmetric biconvex lens made of glass (refractive index $μ_{L} = 1.5$ ) has a focal length of $f_{L} = 20 cm$ in air. It is placed on a horizontal plane mirror. An object is placed on the principal axis at a distance of $30 cm$ from the lens. Match the following properties of the system with their corresponding values.

List - I	List - II
(P) Position of the final image (from the lens) when the lens is in air and placed on the plane mirror.	(1) $7.5 cm$
(Q) Position of the final image (from the lens) if the space between the lens and the plane mirror is filled with water (refractive index $μ_{W} = 4/3$ ).	(2) $10 cm$
(R) Effective power of the lens-mirror system (in Diopters) when the lens is in air and placed on the plane mirror.	(3) $12 cm$
(S) Position of the final image (from the lens) if the biconvex lens is replaced by a plano-convex lens (same material, curved surface radius same as one side of the original biconvex lens) placed on the plane mirror.	(4) $10 Diopters$
	(5) $15 cm$

TestHub · Accepted Answer

The lens-mirror system acts as an equivalent mirror. Its effective focal length is $F_{e ff} = f_{l e n s} /2$ . P: For lens in air, $F_{e ff} = f_{L} /2 = 20/2 = 10 cm$ . Using mirror formula $1/ v + 1/ u = 1/ F_{e ff}$ , with $u = - 30 cm$ : $1/ v + 1/ (- 30) = 1/10 \Rightarrow 1/ v = 1/10 + 1/30 = 4/30 \Rightarrow v = 7.5 cm$ . Q: For a symmetric biconvex lens, $1/ f_{L} = (μ_{L} - 1) (2/ R)$ . So $1/20 = (1.5 - 1) (2/ R) \Rightarrow R = 20 cm$ . When water fills the space, it forms a plano-concave water lens with $R = 20 cm$ . Its focal length is $1/ f_{W} = (μ_{W} - 1) (1/\infty - 1/ R) = (4/3 - 1) (- 1/20) = - 1/60 cm$ . The combined lens has $1/ f_{L}^{'} = 1/ f_{L} + 1/ f_{W} = 1/20 - 1/60 = 2/60 = 1/30 cm$ , so $f_{L}^{'} = 30 cm$ . The effective focal length of the system is $F_{e ff}^{'} = f_{L}^{'} /2 = 30/2 = 15 cm$ . Using mirror formula: $1/ v + 1/ (- 30) = 1/15 \Rightarrow 1/ v = 1/15 + 1/30 = 3/30 \Rightarrow v = 10 cm$ . R: Effective power $P = 1/ F_{e ff}$ (in meters). For the system in air, $F_{e ff} = 10 cm = 0.1 m$ . So $P = 1/0.1 = 10 Diopters$ .

S: A plano-convex lens with $R = 20 cm$ has focal length $1/ f_{pl an o} = (μ_{L} - 1) (1/ R - 1/\infty) = (1.5 - 1) (1/20) = 1/40 cm$ , so $f_{pl an o} = 40 cm$ . The effective focal length of this system is $F_{e ff}^{''} = f_{pl an o} /2 = 40/2 = 20 cm$ . Using mirror formula: $1/ v + 1/ (- 30) = 1/20 \Rightarrow 1/ v = 1/20 + 1/30 = 5/60 \Rightarrow v = 12 cm$ .

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