QJEEM23Ph1c02: JEE Main 2023 Physics - Class 12 JEEM23Ph1c02 Solution: The refractive index of a transparent liquid filled in an…

To find the angle of minimum deviation for the liquid in the equilateral hollow prism, we can use the formula for the angle of minimum deviation based on the refractive index $n$ and the prism angle $A$ :

$n = \frac{\sin \frac{A + \delta_m}{2}}{\sin \frac{A}{2}}$

In this case, the refractive index $n = \sqrt{2}$ , and since the prism is equilateral, the prism angle $A = 60^\circ$ . Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\delta_m$ :

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\sin \frac{60^\circ}{2}}$

First, let's find the sine of half the prism angle:

$\sin \frac{60^\circ}{2} = \sin 30^\circ = \frac{1}{2}$

Now, we can substitute this value into the formula:

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\frac{1}{2}}$

To isolate the sine term, we multiply both sides by $\frac{1}{2}$ :

$\frac{\sqrt{2}}{2} = \sin \frac{60^\circ + \delta_m}{2}$

Now, we can find the angle inside the sine function:

$\frac{60^\circ + \delta_m}{2} = \sin^{-1} \frac{\sqrt{2}}{2} = 45^\circ$

Finally, we can solve for the angle of minimum deviation $\delta_m$ :

$60^\circ + \delta_m = 2 \cdot 45^\circ$

$\delta_m = 90^\circ - 60^\circ = 30^\circ$

The angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\circ$ .