A wide slab consisting of two media of refractive indices n1 and n2 is placed in air as shown in the figure- A ray of light is incident from medium n1 to n2 at an angle - where sin- is slightly larger than 1n1- Take refractive index of air as 1- which of the following statements is are correct-

We have sinθ gt;1n_1 i.e. sinθ_1 gt;1n_1⇒ n_1sinθ_1=n_2sinθ_2 ⇒ sinθ_2=n_1sinθ_1n_2 If n_1=n_2 then, θ_2=θ_1 n_2sinθ_2=(1)sinθ_3⇒ sinθ_3=n_2sinθ_2=n_1sinθ ...

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Byju's Answer

Standard X

Physics

Snell's Law

A wide slab c...

Question

A wide slab consisting of two media of refractive indices $n_{1}$ and $n_{2}$ is placed in air as shown in the figure. A ray of light is incident from medium $n_{1}$ to $n_{2}$ at an angle $θ$ , where $sin θ$ is slightly larger than $\frac{1}{n_{1}}$ . Take refractive index of air as $1$ , which of the following statement(s) is (are) correct?

The light ray is reflected back into the medium of refractive index

n_{1}

n_{2} = 1

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The light ray is finally reflected back into the medium of refractive index

n_{1}

n_{2} < n_{1}

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The light ray is finally reflected back into the medium of refractive index

n_{1}

n_{2} > n_{1}

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The light ray enters air if

n_{2} = n_{1}

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Solution

The correct option is C The light ray is finally reflected back into the medium of refractive index $n_{1}$ if $n_{2} > n_{1}$
We have

$sin θ > \frac{1}{n_{1}}$ i.e. $sin θ_{1} > \frac{1}{n_{1}} \Rightarrow n_{1} sin θ_{1} = n_{2} sin θ_{2}$
$\Rightarrow sin θ_{2} = \frac{n_{1} sin θ_{1}}{n_{2}}$
If $n_{1} = n_{2}$ then, $θ_{2} = θ_{1}$
$n_{2} sin θ_{2} = (1) sin θ_{3} \Rightarrow sin θ_{3} = n_{2} sin θ_{2} = n_{1} sin θ_{1}$
$\Rightarrow sin θ_{1} = \frac{sin θ_{3}}{n_{1}} > \frac{1}{n_{1}} \Rightarrow sin θ_{3} > 1$
$\Rightarrow θ_{3} > 90^{\circ}$
This means ray can't enter air.
For $n_{1} > n_{2}; sin θ_{1} = \frac{n_{2}}{n_{1}} > \frac{1}{n_{1}}$
$\Rightarrow sin θ_{2} > \frac{1}{n_{2}}$
For surface $2 -$ air interface
$n_{2} sin θ_{2} = sin θ_{3} \Rightarrow sin θ_{2} = \frac{sin θ_{3}}{n_{2}} > \frac{1}{n_{2}}$
$θ_{2} > 90^{\circ}$ , it means ray is reflected back in medium of refractive index $n_{2}$

For surface $1 - 2$ inteface
$n_{2} sin θ_{2} = n_{1} sin θ_{1}$
$sin θ_{2 c} = \frac{n_{1}}{n_{2}}$ , where $θ_{2 c}$ is critical angle.
For ray to enter medium of refractive index $n_{1}$
$θ_{2} < θ_{2 c}$
$\Rightarrow sin θ_{2} < sin θ_{2 c}$
$\Rightarrow \frac{n_{1}}{n_{2}} sin θ_{1} < \frac{n_{1}}{n_{2}}$
$\Rightarrow sin θ_{1} < 1 \Rightarrow θ_{1} < 90^{\circ}$ , which is true
Hence ray enters medium of refractive index $n_{1}$ .
For $n_{2} > n_{1}$
$\frac{n_{2}}{n_{1}} sin θ_{2} > \frac{n_{2}}{n_{1}}$
also, $sin θ_{2} > \frac{1}{n_{2}}$
For surface $2 -$ air interface
$n_{2} sin θ_{2} = sin θ_{3}$
$sin θ_{2} = \frac{sin θ_{3}}{n_{2}} > \frac{1}{n_{2}}$
$\Rightarrow θ_{2} > 90^{\circ}$ , it means ray is reflected back in medium of refractive index $n_{2}$ .

$n_{2} sin θ_{2} = n_{1} sin θ_{1}$
$sin θ_{1} = \frac{n_{2}}{n_{1}} sin θ_{2}$
$sin θ_{2 c} = \frac{n_{1}}{n_{2}},$ where, $θ_{2 c}$ is critical angle.
For ray to enter medium of refractive index $n_{1}$
$θ_{2} < θ_{2 c}$
$sin θ_{2} < sin θ_{2 c}$
$\frac{n_{1}}{n_{2}} sin θ_{1} < \frac{n_{1}}{n_{2}}$
$\Rightarrow sin θ_{1} < 1 \Rightarrow θ_{1} < 90^{\circ}$ , which is true
Hence, ray enters medium of refractive index $n_{1}$
Let $n_{2} = 1$

$n_{1} sin θ_{1} = n_{2} sin θ_{2}$
$∵ n_{2} = 1 \Rightarrow n_{1} sin θ_{1} = sin θ_{2}$
$sin θ_{1} = \frac{sin θ_{2}}{n_{1}} > \frac{1}{n_{1}}$
$\Rightarrow sin θ_{2} > 1 \Rightarrow θ_{2} > 90^{\circ}$
Ray is reflected back in the medium of reflective index $n_{1}$

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