At the interface between two materials having refractive indices and , the critical angle for reflection of an em wave is . Then material is replaced by another material having refractive index , such that the critical angle at the interface between an

Question

At the interface between two materials having refractive indices $n_{1}$ and $n_{2}$ , the critical angle for reflection of an em wave is $θ_{1 C}$ . The $n_{2}$ material is replaced by another material having refractive index $n_{3}$ , such that the critical angle at the interface between $n_{1}$ and $n_{3}$ material is $θ_{2 C}$ . If $n_{3} > n_{2} > n_{1}$ ; $\frac{n_{2}}{n_{3}} = \frac{2}{5}$ and $\sin θ_{2 C} - \sin θ_{1 C} = \frac{1}{2}$ , then $θ_{1 C}$ is [2025]

Smart Achievers · Accepted Answer

(none)

$n_{2}$

$n_{3} \sin (θ_{2 C}) = n_{1} \Rightarrow s in (θ_{2 C}) = \frac{n_{1}}{n_{3}}$

Also, $n_{3} = \frac{5 n_{2}}{2} \Rightarrow s in (θ_{2 C}) = \frac{2 n_{1}}{5 n_{2}}$

$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{5}{2} \cdot \sin (θ_{2 C}) \Rightarrow s in (θ_{2 C}) - \sin (θ_{1 C}) = \frac{1}{2}$

Given, $\frac{2 n_{1}}{5 n_{2}} - \frac{n_{1}}{n_{2}} = \frac{1}{2} \Rightarrow \frac{n_{1}}{n_{2}} (\frac{- 3}{5}) = \frac{1}{2}$

Coming out to be (–ve).

Solution

1 $\sin^{- 1} (\frac{1}{6 n_{1}})$

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

3 $\sin^{- 1} (\frac{5}{6 n_{1}})$

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

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Solution

1 sin–1(16n1)

2 sin–1(23n1)

3 sin–1(56n1)

4 sin–1(13n1)

Ans. (none) n2 sin(θ1C)=n1 ⇒ sin(θ1C)=n1n2 n3 sin(θ2C)=n1 ⇒ sin(θ2C)=n1n3 Also, n3 =5n22 ⇒ sin(θ2C)=2n15n2 ⇒ n1n2=52·sin(θ2C) ⇒ sin(θ2C)–sin(θ1C)=12 Given, 2n15n2–n1n2=12 ⇒ n1n2(–35)=12 Coming out to be (–ve).

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1 $\sin^{- 1} (\frac{1}{6 n_{1}})$

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

3 $\sin^{- 1} (\frac{5}{6 n_{1}})$

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