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 $\theta$, where $\sin \theta$ is slightly larger than $\frac{1}{n_1}$. Take refractive index of air as $1$, which of the following statement(s) is (are) correct?

• A. The light ray enters air if $n_2=n_1$
• B. The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2<n_1$
• C. The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2>n_1$
• D. The light ray is reflected back into the medium of refractive index $n_1$ if $n_2=1$

Answer 1

Answer: (B), (C), (D)

The light ray is reflected back into the medium of refractive index $n_1$ if $n_2=1$

We have

$\sin \theta >\frac{1}{n_1}$ i.e. $\sin {\theta }_1>\frac{1}{n_1}\Rightarrow n_1\sin {\theta }_1=n_2\sin {\theta }_2$
$\Rightarrow \sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}$
If $n_1=n_2$ then, ${\theta }_2={\theta }_1$
$n_2\sin {\theta }_2=\left(1\right)\sin {\theta }_3\Rightarrow \sin {\theta }_3=n_2\sin {\theta }_2=n_1\sin {\theta }_1$
$\Rightarrow \sin {\theta }_1=\frac{\sin {\theta }_3}{n_1}>\frac{1}{n_1}\Rightarrow \sin {\theta }_3>1$
$\Rightarrow {\theta }_3>{90}^{\circ }$
This means ray can't enter air.
For $n_1>n_2;\sin {\theta }_1=\frac{n_2}{n_1}>\frac{1}{n_1}$
$\Rightarrow \sin {\theta }_2>\frac{1}{n_2}$
For surface $2-$air interface
$n_2\sin {\theta }_2=\sin {\theta }_3\Rightarrow \sin {\theta }_2=\frac{\sin {\theta }_3}{n_2}>\frac{1}{n_2}$
${\theta }_2>{90}^{\circ }$, it means ray is reflected back in medium of refractive index $n_2$

For surface $1-2$ inteface
$n_2\sin {\theta }_2=n_1\sin {\theta }_1$
$\sin {\theta }_{2c}=\frac{n_1}{n_2}$, where ${\theta }_{2c}$ is critical angle.
For ray to enter medium of refractive index $n_1$
${\theta }_2<{\theta }_{2c}$
$\Rightarrow \sin {\theta }_2<\sin {\theta }_{2c}$
$\Rightarrow \frac{n_1}{n_2}\sin {\theta }_1<\frac{n_1}{n_2}$
$\Rightarrow \sin {\theta }_1<1\Rightarrow {\theta }_1<{90}^{\circ }$, which is true
Hence ray enters medium of refractive index $n_1$.
For $n_2>n_1$
$\frac{n_2}{n_1}\sin {\theta }_2>\frac{n_2}{n_1}$
also, $\sin {\theta }_2>\frac{1}{n_2}$
For surface $2-$air interface
$n_2\sin {\theta }_2=\sin {\theta }_3$
$\sin {\theta }_2=\frac{\sin {\theta }_3}{n_2}>\frac{1}{n_2}$
$\Rightarrow {\theta }_2>{90}^{\circ }$, it means ray is reflected back in medium of refractive index $n_2$.

$n_2\sin {\theta }_2=n_1\sin {\theta }_1$
$\sin {\theta }_1=\frac{n_2}{n_1}\sin {\theta }_2$
$\sin {\theta }_{2c}=\frac{n_1}{n_2},$ where, ${\theta }_{2c}$ is critical angle.
For ray to enter medium of refractive index $n_1$
${\theta }_2<{\theta }_{2c}$
$\sin {\theta }_2<\sin {\theta }_{2c}$
$\frac{n_1}{n_2}\sin {\theta }_1<\frac{n_1}{n_2}$
$\Rightarrow \sin {\theta }_1<1\Rightarrow {\theta }_1<{90}^{\circ }$, which is true
Hence, ray enters medium of refractive index $n_1$
Let $n_2=1$

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$
$\because n_2=1\Rightarrow n_1\sin {\theta }_1=\sin {\theta }_2$
$\sin {\theta }_1=\frac{\sin {\theta }_2}{n_1}>\frac{1}{n_1}$
$\Rightarrow \sin {\theta }_2>1\Rightarrow {\theta }_2>{90}^{\circ }$
Ray is reflected back in the medium of reflective index $n_1$