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A ray of light from a denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of 90^{\circ} with each other. The angles of reflection and refraction are r \& r^{\prime}respectively. The critical angle is

Option: 1

\mathrm{\sin ^{-1}(\tan r)}


Option: 2

\mathrm{\sin ^{-1}(\cot I)}


Option: 3

\mathrm{\tan ^{-1}(\operatorname{Sin} r)}


Option: 4

\mathrm{\tan ^{-1}(\operatorname{Sin} I)}


Answers (1)

best_answer

Applying Snell's Law for refraction,

\frac{\operatorname{Sin} \mathrm{i}}{\operatorname{Sin} \mathrm{r}^{\prime}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\quad \ldots(1)

From the given condition, r+r^{\prime}=90
\Rightarrow \quad \operatorname{Sin} r^{\prime}=\operatorname{Cos} r\quad \ldots(2)

Solution of (1) and (2) yields,
\frac{\sin \mathrm{i}}{\cos \mathrm{r}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\quad \ldots(3)

According to the Law for refraction;
\mathrm{i}=\mathrm{r}\quad \ldots(4)
Using (3) and (4) we obtain
\frac{\operatorname{Sin} \mathrm{i}}{\operatorname{Cos} \mathrm{i}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}
\Rightarrow \quad \tan \mathrm{i}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \quad \ldots(5)

Also,
\sin \theta_{c}=\frac{n_{2}}{n_{1}}$, Using (5) we obtain $\theta_{c}=\sin ^{-1}(\tan i)=\sin ^{-1}(\tan r)

 

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sudhir kumar

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