Question

A particular optical fiber consist of a glass core (index of refraction n₁) surrounded by cladding (index of refraction n₂ < n₁). Suppose a beam of light enters the fibre from air at an angle q with the fibre axis as shown in figure. Show that the greatest possible value of q for which a ray can be propagated down the fibre is given by –

• A.
• B.
• C.
• D. $\theta = \sin ^ { - 1 } \sqrt { 1 - \left( \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right) ^ { 2 } }$

Answer 1

Answer: (A)

Answer 2

r + i = 90

For q max. Ž r should be max.

i will be minimum for TIR minimum value of i

i = c

r = 90 – c …… (i)

sin r = sin(90 – c) = cos c = $\sqrt { 1 - \sin ^ { 2 } \mathrm { c } } = \sqrt { 1 - \left( \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right) ^ { 2 } }$

$\sin \mathrm { r } = \frac { \sqrt { \mathrm { n } _ { 1 } ^ { 2 } - \mathrm { n } _ { 2 } ^ { 2 } } } { \mathrm { n } _ { 1 } }$ ……(ii)

by snell's law

1 × sin q = n₁ sin r = $\sqrt { \mathrm { n } _ { 1 } ^ { 2 } - \mathrm { n } _ { 2 } ^ { 2 } }$

$\theta = \sin ^ { - 1 } \sqrt { \mathrm { n } _ { 1 } ^ { 2 } - \mathrm { n } _ { 2 } ^ { 2 } }$

Option (1) is correct