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Question
JEE Main
Physics

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.

The correct answer is
$45^\circ$

 To find the angle $\theta$ of the prism, we use the following information and concepts:

  1. The refractive index of the material of the prism is given as \(\sqrt{2}\).
  2. The incident ray is parallel to the base of the prism, and the emergent ray grazes along the second surface. This implies that the angle of incidence \(i\) is equal to the angle of minimum deviation \(D_m\) because for grazing emergence, \(\theta_e = 90^\circ\).
  3. The refractive index \(n\) relates to the angle of the prism \(\theta\) and the angle of minimum deviation \(D_m\) via the formula:

\(n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

For grazing emergence, \(A + D_m = 90^\circ\).

  1. Based on this setup:
    1. Given \(n = \sqrt{2}\), let's equate and solve the expression: 
      \(\sqrt{2} = \frac{\sin\left(\frac{A + 90^\circ - A}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

Simplifying, we have:

\(\sqrt{2} = \frac{\sin(45^\circ)}{\sin\left(\frac{A}{2}\right)}\)

Knowing \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\), we substitute:

\(\sqrt{2} = \frac{\frac{1}{\sqrt{2}}}{\sin(\frac{A}{2})}\)

Simplifying, we find:

\(\sin\left(\frac{A}{2}\right) = \frac{1}{2}\)

The angle for which \(\sin(x) = \frac{1}{2}\) is \(x = 30^\circ\).

  1. Therefore, \(\frac{A}{2} = 30^\circ\)

Thus, \(A = 60^\circ\), and by referring to the properties and setup defined, the individual internal angles meet at the indicated \(\theta = 45^\circ\).

Conclusion: Therefore, the angle \(\theta\) of the prism is 45°.

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