Question

A diverging lens with f = –36.5 cm is placed 14.0 cm behind a converging lens with f = 20 cm. Where will an object at infinity be focused?

Answer 1

Solution.

Focal length of diverging lens,

f_d= -33.5 cm

focal length of converging lens,

f_c = 20 cm

Now let s is object distance and s' is image distance.

Now using lens formula for conversing lens

$\frac1{f_c}=\frac1{s}+\frac1{s'}$

Putting all values(object is at s=infinity)

Then image distance

$\frac1{s'}= \frac1{f_c}- \frac1{s}$

$=\frac1{20}- \frac1{\infin}$

$s'=20cm$

Now using lens formula for diversing lens

s''=14-20=-6cm

Using lens formula

$\frac{1}{-33.5} = \frac{1}{-6 }+ \frac{1}{s''' }\\s'''=\frac{(6)\times(-33.5)}{(-33.5+6)}\\s'''=\frac{(-201) }{(-27.5) }$

image distance for diverging lens

s''' = 7.3 cm (behind the diverging lens)