Question

Two plano convex lenses each of focal length 10cm and refractive index 3/2 are placed as shown in the future. in the space left, water (r.i=4/3) is filled. the whole arrangement is in air. the optical power of the system is

Answer 1

for the focal length of lens given that F = 10 cm

now by lens maker's formula

$\frac{1}{f} = (\mu - 1)(\frac{1}{R})$

$\frac{1}{10} = (1.5 - 1)\frac{1}{R}$

R = 5 cm

First of all if a parallel beam of light incident on the flat surface of lens then the lens will converge the light at its focus

this position is given us at 10 cm.

now this image will behave like an object for water surface which is curved by radius R = 5cm

now for curved surface refraction

$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

$\frac{4/3}{v} - \frac{1}{10} = \frac{4/3 - 1}{-5}$

$v = 40 cm$

so whole combination of lens will converge the light at 40 cm

so power of the system is given as

$P = \frac{1}{0.40} = 2.5D$

so power will be 2.5D

Answer 2

f1=10 R=5cm bcoz f1 = R/n-1

Same of f3

And f2 which is filled with water

f2= R/2(n-1) {these are short formula derived from lens maker formula}

So f2= 5/2(4/3-1) = 15/2

By combination of lens

P(cm)= 1/f1+ 1/f3+ 1/f2

P= 1/10+ 1/10+ (-2/15)

P= (3+3-4)/30

P= 2/30 cm

P(diopter) = 100* 2/30

P= 6.67 diopter