Question

A thin convex lens of material of refractive index 5/3 has focal length of 30 cm in air,if it is submerged in a liquid 4/3 then irs focal length will be

Answer 1

This is the correct answer.

Answer 2

Concept:

Lens Maker's formula is given by,

1/f = (n₂ - n₁)/n₁ [1/R₁ - 1/R₂]

where f = focal length

n₂ = refractive index of the material of the lens

n₁ = refractive index of the medium in which the lens is kept

R₁ and R₂ = radius of curvatures of the material of the lens

Given:

Refractive index of the material of the convex lens, $n_g$ = 5/3

Focal length of the lens, $f_g$ = 30 cm

Refractive index of liquid, $n_l$ = 4/3

Find:

The focal length of the lens when it is submerged in the liquid.

Answer:

The focal length of the lens when it is submerged in the liquid is 80 cm.

Solution:

Taking the first case, when the lens is kept in the air.

Refractive index of air, $n_a$ = 1

Applying lens maker's formula, we have

1/$f_g$ = ($n_g - n_a$)/$n_a$ [1/R₁ - 1/R₂]

1/30 = (5/3 - 1)/1 [1/R₁ - 1/R₂]

1/30 = (5-3)/3 [1/R₁ - 1/R₂]

1/30 = 2/3[1/R₁ - 1/R₂]

(1/30)÷(2/3) = [1/R₁ - 1/R₂]

(1/30) × (3/2) = [1/R₁ - 1/R₂]

1/20 = [1/R₁ - 1/R₂]

[1/R₁ - 1/R₂] = 1/20

$[\frac{1}{R_1} - \frac{1}{R_2} ] = \frac{1}{20}$          --------------------------- (i)

Now the lens is submerged in the liquid. Let the focal length of the lens in the liquid be $f_l$.

Applying lens maker's formula, we have

1/ $f_l$ = ($n_g - n_l$)/$n_l$ [1/R₁ - 1/R₂]

1/ $f_l$ = $(\frac{5}{3} - \frac{4}{3} )/(\frac{4}{3} )$ [1/R₁ - 1/R₂]

1/ $f_l$ = (1/3)/(4/3) [1/R₁ - 1/R₂]

1/ $f_l$ = (1/4) [1/R₁ - 1/R₂]

1/ $f_l$ = (1/4) × (1/20)           [From (i)]

1/ $f_l$ = 1/80

$f_l$ = 80 cm

Hence, the focal length of the lens when it is submerged in the liquid is $f_l$ = 80 cm.

#SPJ2