A ray of light refracts from medium 1 into a thin layer of medium 2 , crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. If the angle of incidence of ray is θ, the value of θ is

Question

A ray of light refracts from medium 1 into a thin layer of medium 2 , crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. If the angle of incidence of ray is θ, the value of θ is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/39086ba8bbf7f99e1e465867e6b3934d-.png" /> (A)  sin -1((8/9)) (B)  sin -1((13/18)) (C)  sin -1((13/16)) (D)  sin -1((8/13))

Tardigrade Admin · Accepted Answer

As the light ray incident at critical angle, sin C=(μ3/μ2)=(1.3/1.8) and C=r ∴ sin r=(1.3/1.8) ⇒ sin θ=(1.8/1.6) × (1.3/1.8)=(13/16) θ= sin -1((13/16))

Q. A ray of light refracts from medium $1$ into a thin layer of medium $2$ , crosses the layer and is incident at the critical angle on the interface between the medium $2$ and $3$ as shown in the figure. If the angle of incidence of ray is $θ$ , the value of $θ$ is

$sin^{- 1} (\frac{8}{9})$

$sin^{- 1} (\frac{13}{18})$

$sin^{- 1} (\frac{13}{16})$

$sin^{- 1} (\frac{8}{13})$

As the light ray incident at critical angle,
$sin C = \frac{μ _{3}}{μ _{2}} = \frac{1.3}{1.8}$ and $C = r$
$∴ sin r = \frac{1.3}{1.8}$
$\Rightarrow sin θ = \frac{1.8}{1.6} \times \frac{1.3}{1.8} = \frac{13}{16}$
$θ = sin^{- 1} (\frac{13}{16})$

Q. A ray of light refracts from medium 1 into a thin layer of medium 2 , crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. If the angle of incidence of ray is θ, the value of θ is

sin−1(98​)

sin−1(1813​)

sin−1(1613​)

sin−1(138​)