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PhyMedAdv 2024 • Paper 2

Find Angle Beta for Minimum Deviation in Consecutive Triangular Prisms

Two equilateral-triangular prisms P1P_1 and P2P_2 are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P1P_1 at an angle of incidence θ\theta such that the outgoing ray undergoes minimum deviation in prism P2P_2. If the respective refractive indices of P1P_1 and P2P_2 are 32\sqrt{\frac{3}{2}} and 3\sqrt{3}, then θ=sin⁡−1[32sin⁡(πβ)]\theta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{\beta} \right) \right], where the value of β\beta is ________.

Question Diagram 1
Official Numerical Answer12

Topics & Concepts

Ray Optics and Optical InstrumentsPrismsReflection and Refraction

Step-by-Step Solution

To find the value of β\beta, we trace the path of the light ray through the two prisms P1P_1 and P2P_2.

Step 1: Minimum Deviation in Prism P2P_2

Prism P2P_2 is an equilateral triangle, so its refracting angle is A2=60∘=π3A_2 = 60^\circ = \frac{\pi}{3}. Its refractive index is given as μ2=3\mu_2 = \sqrt{3}.

For a prism undergoing minimum deviation, the angle of refraction r2′r_2' at the first surface is related to the apex angle A2A_2 by: r2′=A22=60∘2=30∘r_2' = \frac{A_2}{2} = \frac{60^\circ}{2} = 30^\circ

Applying Snell's law at the first surface of P2P_2 (transition from vacuum to P2P_2): 1⋅sin⁡(i2′)=μ2sin⁡(r2′)1 \cdot \sin(i_2') = \mu_2 \sin(r_2') sin⁡(i2′)=3⋅sin⁡(30∘)=3⋅12=32\sin(i_2') = \sqrt{3} \cdot \sin(30^\circ) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}

Thus, the angle of incidence on prism P2P_2 is: i2′=60∘i_2' = 60^\circ

Step 2: Emergence Angle from Prism P1P_1

Since the sides of prisms P1P_1 and P2P_2 are parallel to each other, the emergent face (right face) of P1P_1 is parallel to the incident face (left face) of P2P_2.

Because the light ray travels through vacuum between two parallel surfaces, the angle of emergence e1e_1 from P1P_1 is equal to the angle of incidence i2′i_2' on P2P_2: e1=i2′=60∘e_1 = i_2' = 60^\circ

Step 3: Refraction inside Prism P1P_1

Prism P1P_1 is also equilateral, so A1=60∘A_1 = 60^\circ. Its refractive index is μ1=32\mu_1 = \sqrt{\frac{3}{2}}.

Applying Snell's law at the second face (emergent face) of P1P_1: μ1sin⁡(r2)=1⋅sin⁡(e1)\mu_1 \sin(r_2) = 1 \cdot \sin(e_1) 32sin⁡(r2)=sin⁡(60∘)=32\sqrt{\frac{3}{2}} \sin(r_2) = \sin(60^\circ) = \frac{\sqrt{3}}{2} sin⁡(r2)=3/23/2=12\sin(r_2) = \frac{\sqrt{3}/2}{\sqrt{3/2}} = \frac{1}{\sqrt{2}}

Thus, the internal angle of refraction at the second face of P1P_1 is: r2=45∘r_2 = 45^\circ

Inside prism P1P_1, the angles of refraction satisfy: r1+r2=A1r_1 + r_2 = A_1 r1+45∘=60∘  ⟹  r1=15∘=π12r_1 + 45^\circ = 60^\circ \implies r_1 = 15^\circ = \frac{\pi}{12}

Step 4: Angle of Incidence θ\theta on Prism P1P_1

Applying Snell's law at the first face of P1P_1: 1⋅sin⁡(θ)=μ1sin⁡(r1)1 \cdot \sin(\theta) = \mu_1 \sin(r_1) sin⁡(θ)=32sin⁡(15∘)=32sin⁡(π12)\sin(\theta) = \sqrt{\frac{3}{2}} \sin(15^\circ) = \sqrt{\frac{3}{2}} \sin\left(\frac{\pi}{12}\right) θ=sin⁡−1[32sin⁡(π12)]\theta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin\left(\frac{\pi}{12}\right) \right]

Comparing this with the given expression: θ=sin⁡−1[32sin⁡(πβ)]\theta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin\left(\frac{\pi}{\beta}\right) \right]

We get: β=12\beta = 12

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