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Chapter 36: Problem 2896

There is a prism with refractive index equal to \(\sqrt{2}\) and the refracting angle equal to \(30^{\circ} .\) One of the refracting surfaces of the prism is polished. A beam of monochromatic light will retrace its path if its angle of incidence over the refracting surface of the prism is \(\ldots \ldots\) (A) \(45^{\circ}\) (B) \(0^{\circ}\) (C) \(60^{\circ}\) (D) \(30^{\circ}\)

Short Answer

Expert verified
In order to find the angle of incidence for which a beam of monochromatic light will retrace its path after reflection on one of the refracting surfaces of a prism with refractive index equal to √2 and refracting angle equal to 30°, we need to apply Snell's Law and find the critical angle. By considering the condition for total internal reflection, we find the critical angle to be 45°. Therefore, the angle of incidence must be equal to the critical angle, which is also 45°. The correct choice is (A) \(45^{\circ}\).

Step by step solution

01

Understand the problem

We are given a prism with a refractive index equal to √2 and the refracting angle equal to 30°. One of the refracting surfaces is polished, and we are asked to find the angle of incidence for which a beam of monochromatic light will retrace its path after reflection.
02

Apply Snell's Law

When light enters the prism, it refracts and follows Snell's Law. According to Snell's law, when light passes from one medium to another, the product of the refractive index and the sine of the angle of incidence (in medium 1) is equal to the product of the refractive index and the sine of the angle of refraction (in medium 2). Mathematically, it can be expressed as: \(n_1 \sin{i_1} = n_2 \sin{i_2}\). Here, \(n_1\) and \(i_1\) are the refractive index and angle of incidence in medium 1, and \(n_2\) and \(i_2\) are the refractive index and angle of refraction in medium 2.
03

Find the critical angle

To find the critical angle, we need to consider the condition for total internal reflection. For this condition, the angle of refraction should be equal to 90°. Thus, applying Snell's law when the beam of light leaves the prism: \(n_{prism} \sin{i_c} = n_{air} \sin{90^{\circ}}\), where \(i_c\) is the critical angle and \(n_{prism}\) and \(n_{air}\) are the refractive indices of the prism and air, respectively. Since the refractive index of air is approximately 1, we can write: \(\sqrt{2} \sin{i_c} = 1\). Solving for \(i_c\), we get: \(\sin{i_c} = \frac{1}{\sqrt{2}}\). Therefore, \(i_c = 45^{\circ}\).
04

Determine the angle of incidence

In order for the light beam to retrace its path, it needs to internally reflect at the critical angle. Therefore, the angle of incidence must be equal to the critical angle. So, the angle of incidence is \(i = i_c = 45^{\circ}\). Thus, the correct choice is: (A) \(45^{\circ}\).

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Most popular questions from this chapter

An equilateral prism deviates a ray through \(45^{\circ}\) for two angles of incidence differing by \(20^{\circ} .\) What is the \(\mathrm{n}\) of the prism? (A) \(1.467\) (B) \(1.573\) (C) \(1.65\) (D) \(1.5\)

A light ray is incident perpendicular to one face of \(90^{\circ}\) prism and is totally internally reflected at the glass-air interface. If the angle of reflection is \(45^{\circ} .\) We conclude that the refractive index \(\ldots\) (A) \(\mu>\sqrt{2}\) (B) \(\mu>(1 / \sqrt{2})\) (C) \(\mu<\sqrt{2}\) (D) \(\mu<(1 / \sqrt{2})\)

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