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

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