Question

Find out the no of reflection after which light ray will exit from (Given $\sin 40°=0.64$)

• A. $130000$
• B. $57803$
• C. 140000
• D. $150000$

Answer 1

The correct option is B

$57803$

Step 1. Given data:

$\sin 40°=0.64$

$$\begin{gathered} {\mathrm{n}}_1=\mathrm{refractive}\mathrm{index}\mathrm{of}\mathrm{air}=1 \\ {\mathrm{n}}_2=\mathrm{refractive}\mathrm{index}\mathrm{of}2\mathrm{nd}\mathrm{media}=1.31 \end{gathered}$$

Step 2: Formula

$$\begin{gathered} {\mathrm{n}}_1\mathrm{sini}={\mathrm{n}}_2\mathrm{sinr} \\ \mathrm{where}{\mathrm{n}}_1,{\mathrm{n}}_2\mathrm{are}\mathrm{refractive}\mathrm{index}\mathrm{of}\mathrm{two}\mathrm{media} \end{gathered}$$

Step 2. Finding out the number of reflections:

$1\sin 40°=1.31\sin r$

$\Rightarrow$ $0.64=1.31\sin r$

$\Rightarrow$ $\sin r=\frac{0.64}{1.31}=0.49\approx 0.5$

$\Rightarrow$ $r=30°$

In $△ABC$, we have

$\begin{array}{rcl}x& =& \frac{20\mu m}{\tan r}\\ & =& 20\sqrt{3}\mu m\end{array}$

Therefore, $n$(number of reflections) is given by

$\begin{array}{rcl}n& =& \frac{2m}{20\sqrt{3}\mu m}\\ & =& \frac{{10}^5}{\sqrt{3}}\\ & =& 57803\end{array}$

Hence, the correct option is (B).