Question

A concave mirror of focal length $20\text{cm}$ is cut into two parts from the middle and the two parts are moved perpendicularly by a distance $1\text{cm}$ from the previous principal axis AB. Find distance between the images formed by the two parts?

• A. $2\text{cm}$
• B. $4\text{cm}$
• C. $5\text{cm}$
• D. $6\text{cm}$

Answer 1

The correct option is A $2\text{cm}$



Let $P_1{Q}_1=$ principal axis for the mirror $M_1,$​
$P_2{Q}_2=$ principal axis for the mirror $M_2$​ $(I_1=$ Image formed by $M_1,I_2=$ Image formed by $M_2)$
Height of object and image are measured with respect to principal axis of respective mirror.

Applying sign convention, we have,
$f=-20\text{cm},u=-10\text{cm}$

and by Mirror formula
$\frac{1}{-10}+\frac{1}{v}=\frac{1}{-20}⟹v=20\text{cm}$
Magnification of image is
$m=\frac{-v}{u}=\frac{-20}{-10}=2=\frac{h_i}{h_o}⟹h_i=2\text{cm}$

(+ve magnification means image and object are on the same side with respect to principal axis.)

$h_{I_{}1}=2\text{cm}$ below the principal axis $P_1{Q}_1$​

$h_{I_{}2}=2\text{cm}$ below the principal axis $P_2{Q}_2$​
$⟹I_1{I}_2=2\text{cm}$