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A ray of light incident at an angle on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is

A. $7.5^{o}$

B. $5^{o}$

C. $15^{o}$

D. $2.5^{o}$

Answers (1)

The correct option is (a) $7.5^{o}$

As it is a thin prism, the distance between the refracting surfaces is non-accountable and (A), the prism angle is also very small. Since angle (A) is small, it implies that both $r_{1}$ and $r_{2}$ are also small because of the relation $A=r_{1}+r_{2}$ . This also applies to both $i_{1}$ and $i_{2}$ .

According to snell's law, 1 $\sin i_{1}=\mu \; \sin r_{1}\Rightarrow i_{1}=\mu\; r_{1}$

Also, 1 $\sin i_{2}=\mu \; \sin r_{2}\Rightarrow i_{2}=\mu\; r_{2}$

Therefore, deviation, $\delta =\left ( i_{1}-r_{1} \right )+\left ( i_{2}-r_{2} \right )$

$\Rightarrow \; \; \; \; \delta =\left ( i_{1}+i_{2} \right )-\left ( r_{1}+r_{2} \right )=\left ( r_{1}+r_{2} \right )\left ( \mu -1 \right )$

$\Rightarrow \delta =A\left ( \mu -1 \right )$

Since, deviation $\delta =A\left ( \mu -1 \right )A$

$=\left ( 1.5-1 \right )\times 5^{o}=2.5^{o}$

The angle of the prism is $5^{o}$ . The ray emr=erges from refracting face of a prism normally.

Then, $i_{2}=r_{2}=0$

As $A= r_{1}+r_{2}\Rightarrow r_{i}=A\; or\; r_{1}=5^{o}$

But $i_{1}=\mu .r_{1}=\frac{3}{2}\times 5=7.5^{o}$

Posted by

infoexpert23

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