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(a) An object is placed in front of a concave mirror. It is observed that
a virtual image is formed. Draw the ray diagram to show the
image formation and hence derive the mirror equation $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

(b) An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 cm. If the refractive
index of the material of the lens is 1·5, find the position and nature of the image formed.

Answers (1)

a)

From $\Delta \; A'\; B'\; F$ and $M\; F\; P$ by similarity criteria.

$\frac{A'B;}{MP}=\frac{B'F}{FP}$ Or $\frac{A'B'}{AB}=\frac{B'F}{FP}\; \; \;\; (PM=BA)$

Similarly from

$\Delta A'B'P$ and $ABP$

$\frac{B'A}{BA}=\frac{B'P}{BP}$

$\frac{B'F}{FP}=\frac{B'P}{BP}$

$B'F=v+f$

$BP=u$

$\therefore$ $\frac{v+f}{f}=\frac{v}{u}$

$1+\frac{v}{f}=\frac{v}{u}$

Dividing throughout by v and applying sign convention

$\frac{1}{v}-\frac{1}{f}=\frac{-1}{u}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

is the mirror equation.

b)

By lensmakers formula

$\\\frac{1}{f}=(\mu-1)[\frac{1}{R_1}-\frac{1}{R_2}]\\$

For a plano-convex lens

$\\R_1=R\\R_2=\infty$

Therefore

$\frac{1}{f}=(\mu-1)[1/R]$

$f=40cm$

$\\\frac{1}{40}=\frac{1}{v}+\frac{1}{30}\\v=-12cm$

Therefore the image is virtual

Posted by

Safeer PP

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