A cylindrical optical fibre -quarter circular shape- of refractive index n-2 and diameter d-4mm is surrounded by air- A light beam is sent into the fibre along its axis as shown in figure- Then the smallest outer radius R -as shown in figure- which no light escapes during first refraction from curved surface of fibre is-2mm4mm8mm6mm

Question

Toppr Admin · Accepted Answer

For no light to escape during the first refraction,
$θ_{min} > C$
Applying sine on both sides,
$sin (θ_{min}) > sin (C)$
$\frac{R - d}{R} > \frac{1}{n}$
$n (R - d) > R$
$n R - n d > R$
$n R - R > n d$
$R (n - 1) > n d$
$R > \frac{n d}{(n - 1)}$
Substituting the values,
$R > \frac{2 \times 4}{(2 - 1)}$
$R > 8 m m$

For no light to escape during the first refraction,θmin>CApplying sine on both sides,sin(θmin)>sin(C)R−dR>1nn(R−d)>RnR−nd>RnR−R>ndR(n−1)>ndR>nd(n−1)Substituting the values,R>2×4(2−1)R>8mm