The problem involves determining the refractive index and angle of deviation for a light ray passing through a prism. Here’s how to solve the question step-by-step:
We have a prism with an angle of \(45^{\circ}\). A light ray is incident at an angle of \(60^{\circ}\), and it emerges at \(90^{\circ}\) on the other surface. We need to find the refractive index \(\mu\) of the prism's material and the angle of deviation \(\delta\).
The incident angle \(i = 60^{\circ}\) and prism angle \(A = 45^{\circ}\).
Using Snell's Law, at the first surface:
\(\mu = \frac{\sin(i)}{\sin(r_1)}\)
Where \(r_1\) is the angle of refraction at the first surface.
The light ray should satisfy the prism's condition:
\(r_2 = 90^{\circ} - A = 90^{\circ} - 45^{\circ} = 45^{\circ}\)
Using the prism formula, \(A = r_1 + r_2\), substitute the values:
\(45^{\circ} = r_1 + 45^{\circ}\)
Solving gives \(r_1 = 0^{\circ}\).
Using Snell's Law:
\(\mu = \frac{\sin(60^{\circ})}{\sin(0^{\circ}+45^{\circ})} = \frac{\sqrt{3}/2}{1/\sqrt{2}}\)
Calculating gives \(\mu = \sqrt{\frac{3}{2}}\).
The angle of deviation, \(\delta\), is given by:
\(\delta = i + e - A\)
Given that the emergent angle (e) is \(90^{\circ}\),
\(\delta = 60^{\circ} + 45^{\circ} - 45^{\circ} = 15^{\circ}\).
Thus, the refractive index \(\mu\) is \(\sqrt{\frac{3}{2}}\) and the angle of deviation \(\delta\) is \(15^{\circ}\), which corresponds to the correct answer:
\(\mu = \sqrt{\frac{3}{2}},\delta = 15^{\circ}\)