The diagram shows a silvered equiconvex lens. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,$\mu_L = \mu$,$\

Vedclass · Accepted Answer

None

Vedclass · Answer

(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

The figure shows an equiconvex lens of focal length $f$. If the lens is cut along $AB$,the focal length of each half will be

$A$ point-like object is placed at a distance of $1\, m$ in front of a convex lens of focal length $0.5\, m$. $A$ plane mirror is placed at a distance of $2\, m$ behind the lens. Find the position and nature of the final image formed by the system.

$A$ plano-convex lens $(f=20 \,cm)$ is silvered at its plane surface. What will be its new focal length (in $\,cm$)?

An equiconvex lens is cut into two parts along its $(i) \ XOX'$ axis and $(ii) \ YOY'$ axis as shown in the figure. Let $f, f', f''$ be the focal lengths of the original lens,each half in the first case,and each half in the second case,respectively. Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module