QJEEM24Ph7f06: JEE Main 2024 Physics - Class 12 JEEM24Ph7f06 Solution: The refractive index of prism is and the ratio of the angle of…

To find the value of the angle of the prism given the refractive index of the prism $\mu = \sqrt{3}$ and the ratio of the angle of minimum deviation ( $\delta_m$ ) to the angle of the prism ( $A$ ) is 1 (i.e., $\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$ ), we can use the prism formula relating these variables.

The formula that relates the angle of deviation $\delta$ , refractive index $\mu$ , angle of prism $A$ , and angle of minimum deviation $\delta_m$ is given by:

$\mu = \frac{\sin \left(\frac{\delta_m + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ .

Given that $\delta_m = A$ , our formula becomes:

$\sqrt{3} = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Simplifying this, we have:

$\sqrt{3} = \frac{\sin(A)}{\sin \left(\frac{A}{2}\right)}$

Now, recall the trigonometric identity:

$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

By setting $\theta = \frac{A}{2}$ , we get:

$\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$

Substituting back into our equation:

$\sqrt{3} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\sqrt{3} = 2\cos\left(\frac{A}{2}\right)$

Dividing by 2 and solving for $\cos\left(\frac{A}{2}\right)$ :

$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$

This corresponds to an angle $\frac{A}{2}$ of $30^\circ$ since the cosine of 30 degrees is $\frac{\sqrt{3}}{2}$ . Thus:

$\frac{A}{2} = 30^\circ$

So, the angle of the prism $A$ is:

$A = 2 \times 30^\circ = 60^\circ$ .

Therefore, the value of the angle of the prism is $60^\circ$ .