Question

A ray of light is incident at angle of $45^\circ$ on one face of a prism with an equilateral triangular base. If it passes symmetrically through the prism, find the refractive index of the material of the prism.

Hint:

Always memorize standard trigonometric values like $\sin 30^\circ$, $\sin 45^\circ$, and $\sin 60^\circ$. Leaving the answer in surd form ($\sqrt{2}$) is perfectly acceptable and often preferred in physics board exams.

Answer 1

Concept:
• The refractive index ($\mu$) of the material of a prism can be determined if the angle of the prism ($A$) and the angle of minimum deviation ($\delta_m$) are known.
• The relationship is given by the Prism Formula, which is derived from Snell's law applied at the symmetric position.
• The formula is: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$.

Step 1: Identify the required angles
From the problem description and the previous part, we have: Angle of the equilateral prism, $A = 60^\circ$. Angle of minimum deviation, $\delta_m = 30^\circ$.

Step 2: Apply the Prism Formula
Substitute these angles into the refractive index formula: \[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
\[ \mu = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]
\[ \mu = \frac{\sin\left(\frac{90^\circ}{2}\right)}{\sin(30^\circ)} \]
\[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]

Step 3: Evaluate the trigonometric functions
Recall standard trigonometric values: $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ $\sin(30^\circ) = \frac{1}{2}$ Substitute these into the equation: \[ \mu = \frac{\left( \frac{1}{\sqrt{2}} \right)}{\left( \frac{1}{2} \right)} \]
\[ \mu = \frac{1}{\sqrt{2}} \times 2 \]
\[ \mu = \frac{2}{\sqrt{2}} \]
Rationalize the fraction: \[ \mu = \sqrt{2} \]
\[ \mu \approx 1.414 \]

Step 4: Conclusion
The refractive index of the material of the prism is $\sqrt{2}$ or approximately $1.414$.