Question

For an isosceles prism of angle $A$ and refractive index $\mu$, it is found that the angle of minimum deviation ${\delta }_m=A$. Which of the following option(s) is/are correct?

• A. At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1=\left(i_1/2\right)$
• B. For this prism the refractive index $\mu$ and the angle of prism A are related as $A=\frac{1}{2}{\cos }^{-1}\left(\frac{\mu }{2}\right)$
• C. For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1={\sin }^{-1}[\sin A\sqrt{4{\cos }^2\frac{A}{2}-1}-\cos A]$
• D. For the angle of incidence $i_1=A$, the ray inside the prism is parallel to the base of the prism

Answer 1

Answer: (A), (C), (D)

$OptionA$

We know for minimum deviation, $i_1=\frac{A+{\delta }_m}{2}$( ${\delta }_m$ : angle of minimum deviation)

Given ${\delta }_m=A$

Hence $i_1=A$

Now, for minimum deviation condition $r_1=A/2$

Hence $r_1=i_1/2$

$Correct$

$OptionB$

$\mu =\frac{sin\left(i_1\right)}{sin\left(r_1\right)}$

$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{Sin\left(A/2\right)}$

$\mu =\frac{sinA}{sinA/2}$ = $2cosA/2$

$Incorrect$

$OptionC$

Emergence = ${90}^o$

$sin{r}_2=\mu$

$r_2=Si{n}^{-1}\mu$

Also $r_1=A-r_2$

$sin{i}_1=\mu sin{r}_1$

$sin{i}_1=\mu sin(A-r_2)$

$sin{i}_1=\mu (sinAcos{r}_2-cosAsin{r}_2)$

But $\mu sin{r}_2=1$

$sin{i}_1=\mu sinAcos{r}_2-cosA$= $sinA\sqrt{4co{s}^{2}A/2-1}-cosA$

$Correct$