The diagram shows a silvered equiconvex lens. | vedclass

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(D) The power of a lens in a medium is given by $P_L = \frac{1}{f_L} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.Here,$\mu_L = \mu$,$\

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(D) The power of a lens in a medium is given by $P_L = \frac{1}{f_L} = (\frac{\mu_L}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.Here,$\mu_L = \mu$,$\mu_m = 2\mu$,$R_1 = R$,and $R_2 = -R$.So,$P_L = (\frac{\mu}{2\mu} - 1)(\frac{1}{R} - \frac{1}{-R}) = (0.5 - 1)(\frac{2}{R}) = -0.5 	imes \frac{2}{R} = -\frac{1}{R}$.The power of the silvered surface (mirror) is $P_M = -\frac{1}{f_M} = -\frac{2}{R_m}$. Since the mirror is concave with radius $R$,$f_M = -R/2$,so $P_M = -\frac{2}{-R/2} = \frac{4}{R}$.The equivalent power of the system is $P_{eq} = 2P_L + P_M = 2(-\frac{1}{R}) + \frac{4}{R} = \frac{2}{R}$.The system acts as a concave mirror with focal length $f_{eq} = -\frac{1}{P_{eq}} = -\frac{R}{2}$.Since the object is placed in a medium with refractive index $2\mu$,the effective focal length is modified by the ratio of refractive indices. However,for a mirror system,the image formation depends on the reflection. The system acts as a concave mirror of focal length $f = -R/2$. If the object is placed at a distance $u = -30\,cm$,the image will be real and inverted if $u > f$. If $u < f$,it will be virtual and erect. Given the options,the system behaves as a mirror where the image properties depend on the object distance relative to the focal length.

Similar Questions

$A$ concave mirror of focal length $f_1$ is placed at a distance of $d$ from a convex lens of focal length $f_2$. $A$ beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance $d$ must equal:

The distance between a convex lens and a plane mirror is $10 \, cm$. The parallel rays incident on the convex lens after reflection from the mirror form an image at the optical centre of the lens. The focal length of the lens will be.....$cm$.

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