Question

A convex lens of focal length 24 cm ( mew =1.5 )is totally immersed in water ( new = 1.33 ). Then it's focal length in water.

Answer 1

Answer:

solution,by unitary methods

for 1.5 mew the focal length is = 24

for 1 mew the focal length is = 24/1.5

for 1.33 mew the focal length willl be {24×1.33/1.5}=21.28 cm

Answer 2

Answer:

Focal length of convex lens in water is 96 cm .

Explanation:

Given :

Focal length of convex lens in air , f = 24 cm .

Refractive index of lens , $n_l=1.5$ .

Refractive index of water , $n_w=1.5$ .

We know , by lens makers formula :

$\dfrac{1}{f}=(\dfrac{n_l}{n}-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

For air , n = 1 .

So ,  $\dfrac{1}{24}=(\dfrac{1.5}{1}-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})\\\\\dfrac{1}{24}=\dfrac{1}{2}(\dfrac{1}{R_1}-\dfrac{1}{R_2})$     ...... ( 1 )

For water , n = 1.33

$\dfrac{1}{f'}=(\dfrac{1.5}{1.33}-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})\\\\\dfrac{1}{f'}=\dfrac{1}{8}(\dfrac{1}{R_1}-\dfrac{1}{R_2})$            ...... ( 2 )

Dividing equation 1 by 2 we get :

$\dfrac{f'}{24}=\dfrac{8}{2}\\\\f'=96\ cm$

Therefore , focal length of convex lens in water is 96 cm .