Question

An object of height 4 cm is kept to the left of and on the axis of converging lens of focal length 10 cm as shown in figure. A plane mirror is placed inclined at ${45}^{\circ }$ to the lens axis 10 cm to the right of the lens (see figure).Find the position and size of the image formed by combination.

Answer 1

Height $=4cm$

Focal length $=10cm=f$

A plane mirror is placed inclined at $45°$

lens axis$=10cm$

Find the position and size of the image by combination

Solution

Standard this lens equation will be used

$\frac{1}{5}=\frac{1}{u}+\frac{1}{v}f=10cmu=10cm\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}v=30cm$

The formula for the magnification

$M=\frac{h_i}{h_o}=-\frac{v}{u}\frac{h_i}{h_o}=-\frac{v}{u}{h}_i=-h_o\frac{v}{u}$

(where $h_i=$ Height of the image $h_o=$ height of the object $u=$ distance of use object from the centre and $v=$ distance of the object from the centre)

$h_i=-4\frac{30}{15}{h}_i=-8cm$