Question
A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of the prism. Calculate the speed of light in the prism.
✓
Answer
(HOST)$\text{D}_{m} = 2i - \text{A}; \text{D}_{m} = 2 \times \frac{3}{4}\text{A} - \text{A} = \text{A}/2$
$\therefore \text{D}_{m} = \frac{60^\circ}{2} = 30^{\circ}$
$\text{n} = \frac{c_1}{c_2} = \frac{\text{sin}\bigg(\frac{\text{A+D}_m}{2}\bigg)}{\text{sing}\frac{A}{2}}$
$\therefore \text{n} = \frac{c_1}{c_2} = \frac{\text{sin}45^\circ}{\text{sin}30^{\circ}} = \sqrt{2}$
$\text{c}_{2} = \frac{c_1}{\sqrt{2}} = \frac{3\times10^{8}}{\sqrt{}2} \text{ms}^{-1} = 1.5 \sqrt{2}\text{ms}^{-1}$
$\approx2 .12\times10^{8} \text{ms}^{-1}$
Alternate Answer
$\text{r}_{1} + \text{r}_{2} = \text{A}$
At minimum deviation,
$2\text{r} = \text{A}$
or $\text{r} = \frac{A}{2}$
$=30^{\circ}$
Also $\text{i} = \frac{3}{4}\times\text{A} = 45^{\circ}$
$\therefore \text{n} = \frac{c}{c_\circ} = \frac{sin\text{i}}{sin\text{r}}$
$\frac{sin45^\circ}{sin30^\circ} =\sqrt{2}$
$\text{c}_{2} = \frac{\text{c}_{1}}{\sqrt{2}} = \frac{3\times10^{8}\text{ms}^{-1}}{\sqrt{2}}$
$ = 1.5\sqrt{2}\times10^{8}\text{ms}^{-1}$
$\cong2.12 \times10^{8}\text{ms}^{-1}$
$\therefore \text{D}_{m} = \frac{60^\circ}{2} = 30^{\circ}$
$\text{n} = \frac{c_1}{c_2} = \frac{\text{sin}\bigg(\frac{\text{A+D}_m}{2}\bigg)}{\text{sing}\frac{A}{2}}$
$\therefore \text{n} = \frac{c_1}{c_2} = \frac{\text{sin}45^\circ}{\text{sin}30^{\circ}} = \sqrt{2}$
$\text{c}_{2} = \frac{c_1}{\sqrt{2}} = \frac{3\times10^{8}}{\sqrt{}2} \text{ms}^{-1} = 1.5 \sqrt{2}\text{ms}^{-1}$
$\approx2 .12\times10^{8} \text{ms}^{-1}$
Alternate Answer
$\text{r}_{1} + \text{r}_{2} = \text{A}$
At minimum deviation,
$2\text{r} = \text{A}$
or $\text{r} = \frac{A}{2}$
$=30^{\circ}$
Also $\text{i} = \frac{3}{4}\times\text{A} = 45^{\circ}$
$\therefore \text{n} = \frac{c}{c_\circ} = \frac{sin\text{i}}{sin\text{r}}$
$\frac{sin45^\circ}{sin30^\circ} =\sqrt{2}$
$\text{c}_{2} = \frac{\text{c}_{1}}{\sqrt{2}} = \frac{3\times10^{8}\text{ms}^{-1}}{\sqrt{2}}$
$ = 1.5\sqrt{2}\times10^{8}\text{ms}^{-1}$
$\cong2.12 \times10^{8}\text{ms}^{-1}$
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