Match the following properties of a lens system with their corresponding values. List - I List - II (P) Focal length of the lens in air (1) 20 cm (Q) Focal length of one of the plano-convex lenses for

Question

Match the following properties of a lens system with their corresponding values.

List - I	List - II
(P) Focal length of the lens in air	(1) $20 cm$
(Q) Focal length of one of the plano-convex lenses formed if the original lens is cut perpendicular to its principal axis at its optical center	(2) $40 cm$
(R) Focal length of the original lens when immersed in water ( $μ_{w} = 4/3$ )	(3) $80 cm$
(S) Power of the equivalent mirror formed if one surface ( $R_{1} = 20 cm$ ) of the original lens is silvered	(4) $0 D$
	(5) $5 D$

TestHub · Accepted Answer

The lens is a symmetric biconvex lens with $R = 20 cm$ and $μ = 1.5$ . For P: In air, $\frac{1}{f _{P}} = (1.5 - 1) (\frac{1}{20} - \frac{1}{- 20}) = 0.5 (\frac{2}{20}) = \frac{1}{20}$ . So $f_{P} = 20 cm$ . (P)-(1) For Q: A plano-convex lens has $R_{1} = 20 cm$ , $R_{2} = \infty$ . $\frac{1}{f _{Q}} = (1.5 - 1) (\frac{1}{20} - \frac{1}{\infty}) = 0.5 (\frac{1}{20}) = \frac{1}{40}$ . So $f_{Q} = 40 cm$ . (Q)-(2) For R: Immersed in water, $μ_{re l} = \frac{1.5}{4/3} = \frac{9}{8}$ . $\frac{1}{f _{R}} = (\frac{9}{8} - 1) (\frac{1}{20} - \frac{1}{- 20}) = \frac{1}{8} (\frac{2}{20}) = \frac{1}{80}$ . So $f_{R} = 80 cm$ . (R)-(3)

For S: The equivalent mirror has power $P_{e q} = 2 P_{l e n s} + P_{mi rror}$ . Here, the silvered surface is a concave mirror with $f_{m} = - R /2 = - 20/2 = - 10 cm$ . So $P_{mi rror} = \frac{1}{- 0.1 m} = - 10 D$ . The lens power is $P_{l e n s} = \frac{1}{0.2 m} = 5 D$ . Thus, $P_{e q} = 2 (5) + (- 10) = 10 - 10 = 0 D$ . (S)-(4)

Physics - Geometrical Optics Question with Solution | TestHub

Options:

Doubts & Discussion