Back surface of a glass (refractive index $\mathrm{n}$ and thickness $\mathrm{t}$ ) is polished to work as a mirror as shown below. A laser beam falls on it and is partially reflected and refracted at the air-glass interface and fully reflected at the mirror surface respectively. A pattern of discrete spots of light is observed on the screen.
The spacing between the spots on the screen will be

• Correct answer
  A

  $\frac{2 t \cos \theta}{\sqrt{n^{2}-\sin ^{2} \theta}}$
• B

  $\frac{2 t \sin \theta}{\sqrt{n^{2}-\sin ^{2} \theta}}$
• C

  $\frac{2 \mathrm{t} \tan \theta}{\sqrt{\mathrm{n}^{2}-\sin ^{2} \theta}}$
• D

  $\frac{2 \mathrm{t} \sin \theta}{\sqrt{1-\frac{\sin ^{2} \theta}{\mathrm{n}^{2}}}}$

Solution

Therefore $\Delta \mathrm{x}=\mathrm{h}_{1}-\mathrm{h}_{2}$ $$ \begin{array}{l} \tan \theta=\frac{\mathrm{d}}{\mathrm{h}_{1}} \& \tan \theta=\frac{\mathrm{d}-\mathrm{L}}{\mathrm{h}_{2}} \\ \Delta \mathrm{x}=\frac{\mathrm{d}}{\tan \theta}-\frac{\mathrm{d}-\mathrm{L}}{\tan \theta} \\ \Delta \mathrm{x}=\frac{\mathrm{L}}{\tan \theta} \end{array} $$ Snell's Law $$ \begin{array}{l} 1 \cdot \sin \theta=\mathrm{n} \cdot \sin \mathrm{r} \Rightarrow 1 \sin \mathrm{r}=\frac{\sin \theta}{\mathrm{n}} \\ \tan \mathrm{r}=\frac{\mathrm{L}}{2 . \mathrm{t}} \Rightarrow \mathrm{L}=2 \mathrm{t} \tan \mathrm{r} \end{array} $$ $$ \mathrm{L}=\frac{2 \mathrm{t} \cdot \sin \theta}{\sqrt{\mathrm{n}^{2}-\sin ^{2} \theta}} $$ $$ \begin{array}{l} \therefore \Delta \mathrm{x}=\frac{2 \mathrm{t} \sin \theta}{\sqrt{\mathrm{n}^{2}-\sin ^{2} \theta} \tan \theta} \\ \tan \theta=\frac{2 \mathrm{t} \cos \theta}{\sqrt{\mathrm{n}^{2}-\sin ^{2} \theta}} \end{array} $$

Hence, the correct answer is option: 'A'.