Question 79: Ray Optics

Question

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

Accepted Answer

The refractive index $n$ of a prism with refracting angle $A$ and angle of minimum deviation $\delta_m$ is given by the formula:
$$ \mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}ight)}{\sin \left(\frac{\mathrm{A}}{2}ight)} $$
According to question the angle of minimum deviation is equal to the refracting angle:
$$ \delta_{\mathrm{m}}=\mathrm{A} $$
Putting in the formula,
$$ \mathrm{n}=\frac{\sin \left(\frac{A+A}{2}ight)}{\sin \left(\frac{A}{2}ight)}=\frac{\sin (A)}{\sin \left(\frac{A}{2}ight)} $$
$$\Rightarrow $$ $\mathrm{n}=\frac{2 \sin \left(\frac{\mathrm{~A}}{2}ight) \cos \left(\frac{\mathrm{A}}{2}ight)}{\sin \left(\frac{\mathrm{A}}{2}ight)}=2 \cos \left(\frac{\mathrm{~A}}{2}ight)$
The value of the refractive index depends on the possible values for the refracting angle $A$.
As A approaches $0^{\circ}, \cos \left(\frac{\mathrm{A}}{2}ight) ightarrow \cos \left(0^{\circ}ight)=1$.
Therefore, n approaches $2 	imes 1=2$.
For minimum deviation to occur, the internal angle of incidence $\mathrm{r}=\frac{\mathrm{A}}{2}$ must be less than the critical angle C to let the light emerge from prism.
$$ \sin \left(\frac{A}{2}ight)<\sin (C)=\frac{1}{n} $$
Using $\mathrm{n}=2 \cos \left(\frac{\mathrm{~A}}{2}ight)$,
$$ 2 \sin \left(\frac{\mathrm{~A}}{2}ight) \cos \left(\frac{\mathrm{A}}{2}ight)<1 \Rightarrow \sin (\mathrm{~A})<1 $$
This condition is satisfied for all $\mathrm{A}<90^{\circ}$.
If $\mathrm{A}=90^{\circ}$, then $\mathrm{n}=2 \cos \left(45^{\circ}ight)=2 	imes \frac{1}{\sqrt{2}}=\sqrt{2}$
For a practical transparent prism where $0<\mathrm{A}<90^{\circ}$, the refractive index must fall in the range:
$$ \sqrt{2}<\mathrm{n}<2 $$
Hence, the correct option is (B).

Answer

$n \geq 2$

Answer

$\sqrt{2} < n < 2$

Answer

$1 < n < 2$

Answer

$\sqrt{2}<n<2\sqrt{2}$

Ray Optics

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

Choose the correct answer: