Ray Optics

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52 of 295

Question 52 of 295

One side of an equilateral prism is painted by a transparent material of refractive index n2n_2. The refractive index of prism is 1.61.6. The minimum value of n2n_2 required for total internal reflection from painted face is _______.

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Answer: (D) 43/54\sqrt{3}/5

Given that the prism is an equilateral triangle, all its refracting angles are 60∘60^\circ.

The incident ray falls normally on one of the faces of the prism. Therefore, the angle of incidence is 0∘0^\circ and the ray enters the prism undeviated.

From the geometry of the figure, the ray strikes the bottom painted face. In the right-angled triangle formed by the ray and the two faces of the prism, the angle between the left face and the bottom face is 60∘60^\circ.

The angle the refracted ray makes with the bottom face is 90∘−60∘=30∘90^\circ - 60^\circ = 30^\circ.

The angle of incidence ii at the bottom painted face is the angle made with the normal to that face:
i=90∘−30∘=60∘i = 90^\circ - 30^\circ = 60^\circ

For total internal reflection to occur at the painted face, the angle of incidence must be greater than or equal to the critical angle θc\theta_c for the prism-paint interface:
i≥θci \ge \theta_c
sin⁡i≥sin⁡θc\sin i \ge \sin \theta_c

The critical angle is given by sin⁡θc=n2n1\sin \theta_c = \dfrac{n_2}{n_1}, where n1=1.6n_1 = 1.6 is the refractive index of the prism and n2n_2 is the refractive index of the paint.

Substituting the values:
sin⁡60∘≥n21.6\sin 60^\circ \ge \dfrac{n_2}{1.6}
32≥n21.6\dfrac{\sqrt{3}}{2} \ge \dfrac{n_2}{1.6}
n2≤1.6×32n_2 \le 1.6 \times \dfrac{\sqrt{3}}{2}
n2≤0.83=435n_2 \le 0.8\sqrt{3} = \dfrac{4\sqrt{3}}{5}

The limiting value of n2n_2 required for total internal reflection is 435\dfrac{4\sqrt{3}}{5}.

Answer: 43/54\sqrt{3}/5

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