A ray of light is incident at an angle i from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively r r prime then the critical angle will be:

Question

A ray of light is incident at an angle i from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively r r prime then the critical angle will be: (A)  sin -1( sin r) (B)  sin -1( tan r prime) (C)  sin -1( tan i) (D)  sin -1( sin i)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/cda4f69664f76b38a9cf97bbdba54c55-.png" /> D μ R =( sin i / sin r ') ⇒ R μ D =( sin r '/ sin i ) (1/ sin c )=( sin r '/ sin i ) sin C =( sin i / sin (90- r ))=( sin i / cos r )( angle i = angle r ) ∴ sin C = tan i C = sin -1( tan i )

Q. A ray of light is incident at an angle $i$ from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively $r & r^{'}$ then the critical angle will be:

$sin^{- 1} (sin r)$

$sin^{- 1} (tan r^{'})$

$sin^{- 1} (tan i)$

$sin^{- 1} (sin i)$

$^{D} μ_{R} = \frac{s i n i}{s i n r ^{'}}$
$\Rightarrow^{R} μ_{D} = \frac{s i n r ^{'}}{s i n i}$
$\frac{1}{s i n c} = \frac{s i n r ^{'}}{s i n i}$
$sin C = \frac{s i n i}{s i n ( 90 - r )} = \frac{s i n i}{c o s r} (∠ i = ∠ r)$
$∴ sin C = tan i$
$C = sin^{- 1} (tan i)$

Q. A ray of light is incident at an angle i from denser to rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively r&r′ then the critical angle will be:

sin−1(sinr)

sin−1(tanr′)

sin−1(tani)

sin−1(sini)