Question

An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.

Answer 1

I hope it helps you :)..

Answer 2

Explanation:

(a) For a converging lens,

The object position,  $u=-20 \mathrm{cm}.$

Focal length, $f=15 \mathrm{cm}.$

Using lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}.$

So,  $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-20}=\frac{-4-3}{60}=-\frac{7}{60}.$

Therefore, v $=-8.57 \mathrm{cm}.$

And magnification, $\mathrm{m}=\frac{v}{u}=\frac{-8.57}{-20}=+0.42.$

(b) For a diverging lens,

The object position, $u=-20 \mathrm{cm}$

Focal length, $f=-15 \mathrm{cm}$

Using lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\text { So, } \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-15}-\frac{1}{-20}=\frac{-4+3}{60}=-\frac{1}{60}.$

Therefore, v= 60 cm

$\text { And magnification, } m=\frac{v}{u}=\frac{60}{-20}=-3.$