A ray of light passes symmetrically through an equilateral glass prism of refractive index… — JAMB UTME

A ray of light passes symmetrically through an equilateral glass prism of refractive index $\sqrt{2}$ . What is the angle of minimum deviation of the ray, in degrees?

Answer: 30 degrees

Answer

The angle of minimum deviation of the ray is

30^\circ

For an equilateral triangular prism, the apex angle

A

60^\circ

. At minimum deviation, light travels symmetrically through the prism and obeys the exact relation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

. Substituting

n = \sqrt{2}

and

A = 60^\circ

gives

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin(45^\circ)

. Equating arguments yields

\frac{60^\circ + D_m}{2} = 45^\circ

, leading directly to

D_m = 30^\circ

Step-by-Step Solution

Determine the refracting angle of the prism

A = 60^\circ

An equilateral prism has interior angles of

60^\circ

each, so the apex angle

A = 60^\circ

Set up the prism formula for minimum deviation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

When light passes symmetrically through a prism, the deviation is at its minimum value

D_m

Substitute known values into the equation

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Given

n = \sqrt{2}

and

A = 60^\circ

, with

\sin(30^\circ) = 0.5

Calculate the sine of the half-angle

\sin\left(\frac{60^\circ + D_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

Multiplying both sides by

\sin(30^\circ) = 0.5

Solve for the minimum deviation angle

D_m

D_m = 30^\circ

Since

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

, we set

\frac{60^\circ + D_m}{2} = 45^\circ \implies 60^\circ + D_m = 90^\circ \implies D_m = 30^\circ

Key Concept

Refraction of light through a prism at the angle of minimum deviation

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