Question

JEE mains Previous year question
chapter :- Geometrical Optics​​

Answer 1

Explanation:

• According to lens lens maker's formula,1/f = (μ - 1) (1/R₁ - 1)/R₂ ) when the lens in the air

• 1/f = ( 3/2 - 1) (1/R₁ - 1/R₂)

• 1/f = 1/2x => f = 2x

• Here, 1/x = 1/R₁ - 1/R₂

In vade of liquid ,where refractive indices are 4/3 and 5/3 ,we get

focal length in first liquid ,

1/f₁ = (μₛ / μₗ₁ - 1) (1/R₁ - 1/R₂ )

1/f₁ = (3/2/4/3 - 1 ) 1/x

• → F₁ is positive .

Nature of lens is not change.

• 1/f₁ = 1/8x = 1/4(2x) = 1/4f → f₁ = 4f

→ focal length in second liquid

• 1/f₂ = (μₛ/μₗ₂ - 1) (1/R₁ - 1/R₂

• 1/f₂ = (3/2/5/3 - 1) (1/x)

• f₂ is negative

nature of lens change i.e . , convex behave as concave.

Answer 2

Answer:

Given, refractive index of convex lens, μ = 3/2

and focal length = f

Let f₁ be the focal length when lens is placed in liquid of μ₁ = 4/3

and f₂ be the focal length when lens is placed in liquid of μ₂ = 5/3.

Now,

using Lens maker's formula:-

1/f = (μ - 1)(1/R₁ - 1/R₂) (in air)

→ 1/f = (3/2 - 1)(1/R₁ - 1/R₂)

→ 1/f = 1/2 (1/R₁ - 1/R₂)

let (1/R₁ - 1/R₂) = 1/R

$→\frac{1}{f} = \frac{1}{2R}$

∴ f = 2R

similarly, (in liquid of μ₁ = 4/3)

1/f₁ = (μ/μ₁ - 1)(1/R₁ - 1/R₂)

$= (\frac{ \frac{3}{2} }{ \frac{4}{3} } - 1) \frac{1}{R}$

$= (\frac{9}{8} - 1) \frac{1}{R}$

$= \frac{1}{8R}$

→f₁ = 8R

→f₁ = 4×2R

→f₁ = 4f

so, f₁ > f

similarly, (in liquid of μ₂ = 5/3)

1/f₂ = (μ/μ₂ - 1)(1/R₁ - 1/R₂)

$= (\frac{ \frac{3}{2} }{ \frac{5}{3} } - 1) \frac{1}{R}$

$= ( \frac{9}{10} - 1) \frac{1}{R}$

$= \frac{ - 1}{10R}$

→ f₂ = -10R

so, f₂ becomes negative.

Hence, (d) f₁ > f and f₂ becomes negative.