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  • An equilateral prism has an angle of deviation  when the angle of incidence is <img a

An equilateral prism has an angle of deviation 30^{\circ} when the angle of incidence is 60^{\circ}. The angle of deviation if a ray is incident normally on a surface is:

Option: 1

50^{\circ}


Option: 2

60^{\circ}


Option: 3

90^{\circ}


Option: 4

40^{\circ}


Answers (1)

best_answer

For a prism, angle of deviation

\mathrm{\delta=I_{1}+I_{2}-A\quad \ldots(1)}
\mathrm{and \, r_{1}+r_{2}=A\quad \ldots(2)}
\mathrm{Here \, \delta=30^{\circ}, I_{1}=60^{\circ}, A=60^{\circ}}

Putting these values in (1) and (2), we get

\mathrm{\mathrm{I}_{2}=30^{\circ} \text { and } \mathrm{r}_{1}+\mathrm{r}_{2}=60^{\circ}}

Also from Snell’s Law
\mathrm{\mu=\frac{\sin l_{1}}{\sin r_{1}}=\frac{\sin l_{2}}{\sin r_{2}} \Rightarrow \frac{\sin 60^{\circ}}{\sin r_{1}}=\frac{\sin 30^{\circ}}{\sin \left(60^{\circ}-r_{1}\right)}}
Solving we get, \mathrm{\tan r_{1}=\frac{3}{2+\sqrt{3}}}
\mathrm{\Rightarrow \sin r_{1}=\frac{3}{2 \sqrt{4+\sqrt{3}}}}

\mathrm{Hence \, \mu=\frac{\sin 60^{\circ}}{\sin r_{1}^{\circ}} \approx 1.383}

Now if the ray is incident normally on this prism angle of incidence at the

\mathrm{\text{Second surface} =60^{\circ} (as\: shown)}

\mathrm{But, \sin 60^{\circ}>\frac{1}{\mu}}
\therefore  Total internal reflection will take place
At the third surface this again incident normally (as shown).

Hence the ray suffer deviation only at the second surface, which is equal to \mathrm{180^{\circ}-2 \times 60^{\circ}=60^{\circ}}.
 

 

Posted by

Kuldeep Maurya

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