Light from a denser medium 1 passes to a rarer medium 2 . When the angle of incidence is θ the partially reflected and refracted rays are mutually perpendicular. The critical angle will be

Question

Light from a denser medium 1 passes to a rarer medium 2 . When the angle of incidence is θ the partially reflected and refracted rays are mutually perpendicular. The critical angle will be (A)  sin -1( cot θ) (B)  sin -1( tan θ) (C)  sin -1( cos θ) (D)  sin -1( sec θ)

Tardigrade Admin · Accepted Answer

μ1 sin θ=μ2 × sin (90°-θ) ⇒ (μ1/μ2)= tan θ For θc ⇒ μ1 × sin θC=μ2 × sin (90°) sin θC=(μ1/μ2)= tan θ ⇒ θC= sin -1( tan θ)

Q. Light from a denser medium $1$ passes to a rarer medium $2.$ When the angle of incidence is $θ$ the partially reflected and refracted rays are mutually perpendicular. The critical angle will be

$sin^{- 1} (cot θ)$

$sin^{- 1} (tan θ)$

$sin^{- 1} (cos θ)$

$sin^{- 1} (sec θ)$

$μ_{1} sin θ = μ_{2} \times sin (9 0^{\circ} - θ)$
$\Rightarrow \frac{μ _{1}}{μ _{2}} = tan θ$
For $θ_{c}$
$\Rightarrow μ_{1} \times sin θ_{C} = μ_{2} \times sin (9 0^{\circ})$
$sin θ_{C} = \frac{μ _{1}}{μ _{2}} = tan θ$
$\Rightarrow θ_{C} = sin^{- 1} (tan θ)$

Q. Light from a denser medium 1 passes to a rarer medium 2. When the angle of incidence is θ the partially reflected and refracted rays are mutually perpendicular. The critical angle will be

sin−1(cotθ)

sin−1(tanθ)

sin−1(cosθ)

sin−1(secθ)