Light travels in two media A and B with speeds 1.8 × 108 ms-1 and 2.4 × 108 ms-1 respectively. Then the critical angle between them is

Question

Light travels in two media A and B with speeds 1.8 × 108 ms-1 and 2.4 × 108 ms-1 respectively. Then the critical angle between them is (A)  sin-1 ((2/3)) (B)  tan-1 ((3/4)) (C)  tan-1 ((2/3)) (D)  sin-1 ((3/4))

Tardigrade Admin · Accepted Answer

We have C= sin-1 ((v1/v2)) = sin-1 ((1.8 × 108/2.4 × 108))=sin-1 ((3/4))

Q. Light travels in two media $A$ and $B$ with speeds $1.8 \times 1 0^{8} m s^{- 1}$ and $2.4 \times 1 0^{8} m s^{- 1}$ respectively. Then the critical angle between them is

$sin^{- 1} (\frac{2}{3})$

$tan^{- 1} (\frac{3}{4})$

$tan^{- 1} (\frac{2}{3})$

$sin^{- 1} (\frac{3}{4})$

$sin^{- 1} (\frac{1}{4})$

We have
$C = s i n^{- 1} (\frac{v _{1}}{v _{2}})$
$= s i n^{- 1} (\frac{1.8 \times 1 0 ^{8}}{2.4 \times 1 0 ^{8}}) = s i n^{- 1} (\frac{3}{4})$

Q. Light travels in two media A and B with speeds 1.8×108ms−1 and 2.4×108ms−1 respectively. Then the critical angle between them is

sin−1(32​)

tan−1(43​)

tan−1(32​)

sin−1(43​)

sin−1(41​)

We have C=sin−1(v2​v1​​) =sin−1(2.4×1081.8×108​)=sin−1(43​)