Light from a denser medium I goes to a rarer medium II . When angle of incidence is θ , the reflected and refracted rays are perpendicular to each other. The critical angle is :

Question

Light from a denser medium I goes to a rarer medium II . When angle of incidence is θ , the reflected and refracted rays are perpendicular to each other. The critical angle is : (A)  sin -1( cos θ )  (B)  sin -1( cot θ )  (C)  sin -1( tan θ )  (D)  sin -1(1)

Tardigrade Admin · Accepted Answer

From laws of reflection, we know that angle of incidence is equal to angle of reflection.

Let

r

be the angle of refraction, along line

A OB

, we have

θ + 9 0^{\circ} + r = 18 0^{\circ}

\Rightarrow r = 90 - θ^{\circ}

Also from Snell's law

_{1} μ_{2} = \frac{s i n i}{s i n r}

In this case

i = θ, r = 9 0^{\circ} - θ

_{1} μ_{2} = \frac{s i n θ}{s i n ( 9 0 ^{\circ} - θ )} = \frac{s i n θ}{c o s θ} = tan θ

Also from definition of critical angle, we have

sin C = \frac{1}{_{2} μ _{1}}

sin C = tan θ

\Rightarrow C = sin^{- 1} (tan θ)

Q. Light from a denser medium $I$ goes to a rarer medium $II$ . When angle of incidence is $θ$ , the reflected and refracted rays are perpendicular to each other. The critical angle is :

$sin^{- 1} (cos θ)$

$sin^{- 1} (cot θ)$

$sin^{- 1} (tan θ)$

$sin^{- 1} (1)$

Q. Light from a denser medium I goes to a rarer medium II . When angle of incidence is θ , the reflected and refracted rays are perpendicular to each other. The critical angle is :

sin−1(cosθ)

sin−1(cotθ)

sin−1(tanθ)

sin−1(1)

Q. Light from a denser medium $I$ goes to a rarer medium $II$ . When angle of incidence is $θ$ , the reflected and refracted rays are perpendicular to each other. The critical angle is :

$sin^{- 1} (cos θ)$

$sin^{- 1} (cot θ)$

$sin^{- 1} (tan θ)$

$sin^{- 1} (1)$