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Chapter 16: Problem 2274

A ray of light is incident normally on one of the faces of a solid prism of apex angle \(30^{\circ}\) and refractive index \(\sqrt{2}\). The angle of minimum deviation is (A) \(39^{\circ}\) (B) \(42^{\circ}\)

Short Answer

Expert verified
The angle of minimum deviation is \(|D| = 30^{\circ}\). However, none of the given options match this value, which could indicate an error in the problem statement or a lack of correct answers among the options.

Step by step solution

01

Recall Snell's Law and the formula for the angle of minimum deviation in a prism

Snell's Law states that \[n_1 \sin\theta_1 = n_2 \sin\theta_2\] Where \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively. The formula for the angle of minimum deviation in a prism is given by \[D = i + r - A\] Where \(D\) is the angle of minimum deviation, \(i\) and \(r\) are the angles of incidence and refraction inside the prism respectively, and \(A\) is the apex angle of the prism.
02

Apply Snell's Law to determine the angle of refraction inside the prism

From the problem, we are given that the light ray is incident normally on the surface of the prism, so the angle of incidence (\(\theta_1\)) is \(0^{\circ}\). We are also given that the refractive index of the prism is \(\sqrt{2}\). Applying Snell's Law, we get \[n_1 \sin\theta_1 = n_2 \sin\theta_2\] Since the angle of incidence is \(0^{\circ}\), \(\sin 0^{\circ} = 0\). So the equation becomes \[0 = \sqrt{2} \sin\theta_2\] This means that \(\sin\theta_2 = 0\), which implies that the angle of refraction inside the prism, (\(\theta_2\)), is also \(0^{\circ}\).
03

Calculate the angle of minimum deviation

From the formula of minimum deviation, we have \[D = i + r - A\] We already have the values of \(i\), \(r\), and \(A\). Substituting them, we get: \[D = 0^{\circ} + 0^{\circ} - 30^{\circ} = -30^{\circ}\] However, since the angle of minimum deviation must be positive, we should take the absolute value of the result. So, the angle of minimum deviation is \(|D| = 30^{\circ}\). Among the given options, none of them match our calculated angle of minimum deviation, so either there is an error in the problem statement or there is no correct answer among the options.

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Most popular questions from this chapter

A Sound wave travels from air to water. The angle of incidence is \(\alpha_{1}\) and the angle of reflection is \(\alpha_{2}\) If the snell's Law is valid then, (A) \(\alpha_{1} \geq \alpha_{2}\) (B) \(\alpha_{1}=\alpha_{2}\) (C) \(\alpha_{1}>\alpha_{2}\) (D) \(\alpha_{1}<\alpha_{2}\)

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