A student studies a glass prism whose refracting angle is 60° and whose refractive index is 1.50…

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A student studies a glass prism whose refracting angle is $60^\circ$ and whose refractive index is $1.50$ . The figure shows a narrow beam of light in air that meets the first face at an angle of incidence of $45^\circ$ . Use $1.00$ for air. Light that leaves the second face makes an angle of emergence with the normal there. Later, the student dips the whole prism in a clear liquid of refractive index $1.20$ and repeats the work at minimum deviation, the position where the angle of deviation is smallest.

A glass prism seen from the side, shaped like an equilateral triangle with its apex at the top and the apex angle marked 60 degrees. A ray in air comes in from the lower left and meets the left face. The normal at that point is dashed, and the angle of incidence is marked 45 degrees. Inside the glass, marked refractive index 1.50, the ray runs across towards the right face. At the right face a dashed normal and a question mark show where the emergent ray will leave.

What is the angle of refraction at the first face?
What is the angle of deviation of the beam in the figure?
What is the minimum deviation of this prism in air?
What is the minimum deviation of the prism in the liquid?
In air, rays with a small angle of incidence reach the second face at more than the critical angle and cannot leave it. What is the smallest angle of incidence at the first face for which a ray does emerge?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Follow the ray through the two faces with Snell's law. Then use the minimum deviation formula, with the relative refractive index when the prism is in the liquid.

(a) First face. $\sin r_1 = \dfrac{\sin 45^\circ}{1.50} = 0.4714$ , so $r_1 = 28.1^\circ$ .

(b) Deviation. At the second face, $r_2 = 60^\circ - 28.1^\circ = 31.9^\circ$ . Then $\sin e = 1.50 \sin 31.9^\circ = 0.7921$ , so $e = 52.4^\circ$ . The deviation is $\delta = 45^\circ + 52.4^\circ - 60^\circ = 37.4^\circ$ .

(c) Minimum deviation in air. $\sin\dfrac{A + \delta_m}{2} = 1.50 \sin 30^\circ = 0.7500$ . So $\dfrac{A + \delta_m}{2} = 48.59^\circ$ and $\delta_m = 37.2^\circ$ .

(d) Minimum deviation in the liquid. The relative index is $\dfrac{1.50}{1.20} = 1.250$ . Then $\sin\dfrac{A + \delta_m}{2} = 1.250 \times \sin 30^\circ = 0.6250$ , so $\dfrac{A + \delta_m}{2} = 38.68^\circ$ and $\delta_m = 17.4^\circ$ .

(e) Smallest angle of incidence. The critical angle of the glass is $\sin C = \dfrac{1.00}{1.50}$ , so $C = 41.8^\circ$ . A ray just emerges when $r_2 = C$ , so $r_1 = 60^\circ - 41.8^\circ = 18.2^\circ$ . Then $\sin i = 1.50 \sin 18.2^\circ$ , which gives $i = 27.9^\circ$ .

Answer: (a) $28.1^\circ$ . (b) $37.4^\circ$ . (c) $37.2^\circ$ . (d) $17.4^\circ$ . (e) $27.9^\circ$ .

Check. Scan the angle of incidence from $20^\circ$ to $90^\circ$ in steps of $0.01^\circ$ . The least deviation found is $37.2^\circ$ in air and $17.4^\circ$ in the liquid, and the first ray that emerges in air is at $27.9^\circ$ .

Watch out. The deviation in (b) is only $0.2^\circ$ above the minimum, because $45^\circ$ is close to the best angle. In the liquid the deviation falls a lot, from $37.2^\circ$ to $17.4^\circ$ . Using $1.50$ instead of the relative index $1.250$ in (d) gives the air value again.

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