Question

IIT JEE maine Previous year question Optics​

Answer 1

Explanation:

let the number of reflections be n.

d = 20m = AB

l = 2m

θ₁ = 40°

OA = x

refractive index of air, μ₁ = 1

refractive index of μ₂ = 1.31

Applying Snell's law at O :-

→ μ₁sini = μ₂sinr

→ sin40° = 1.31 sinθ₂

→ sinθ₂

$→sinθ₂ = \frac{ \sin40° }{1.31}$

$→sinθ₂ = \frac{0.64}{1.31}$

$→sinθ₂≈ \frac{1}{2}$

→ θ₂ = 30°

Now, In Δ OAB,

$\tan(θ₂) = \frac{d}{x}$

$→x = \frac{20μm}{ \tan(30°) }$

→ x = 20√3 μm

The reflection will occur at every x distance in the optical fibre.

→ number of reflections × x = total length of the optical fibre.

→ nx = l

$→ n = \frac{l}{x}$

$→ n = \frac{2 \: m}{20 \sqrt{3} \times {10}^{ - 6} \: m }$

$→ n = 0.57 \times {10}^{ 5}$

∴ n = 57000