JEE mains Previous year question
chapter :- Geometrical Optics
Answers
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:
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
∴ f = 2R
similarly, (in liquid of μ₁ = 4/3)
1/f₁ = (μ/μ₁ - 1)(1/R₁ - 1/R₂)
→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)
→ f₂ = -10R
so, f₂ becomes negative.
Hence, (d) f₁ > f and f₂ becomes negative.