A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is :
![a right angled prism at point P with the angle of incidence]()
- \(\frac{\sqrt5}{4}\)
- \(\frac{\sqrt5}{2}\)
- \(\frac{\sqrt3}{4}\)
- \(\frac{\sqrt3}{2}\)
The Correct Option is B
Solution and Explanation
Step 1: Prism Relation Analysis
Given the prism relation:
$$ r_1 + c = A $$
Isolate \( r_1 \):
$$ r_1 = 90^\circ - c \quad \text{...(1)} $$
Step 2: Derivation of \( \cos c \)
From the given information:
$$ \sin c = \frac{1}{\mu} $$
Applying the Pythagorean trigonometric identity:
$$ \cos c = \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Step 3: Application of Snell's Law at the Incidence Surface
Snell's law at the first surface yields:
$$ \sin 30^\circ = \mu \sin (r_1) $$
Substitute \( r_1 = 90^\circ - c \):
$$ \frac{1}{2} = \mu \sin (90^\circ - c) $$
Utilizing the identity \( \sin (90^\circ - c) = \cos c \):
$$ \frac{1}{2} = \mu \times \frac{\sqrt{\mu^2 - 1}}{\mu} $$
Simplify the equation:
$$ \frac{1}{2} = \frac{\sqrt{\mu^2 - 1}}{1} $$
Step 4: Determination of \( \mu \)
Square both sides of the equation:
$$ \frac{1}{4} = \mu^2 - 1 $$
Rearrange to solve for \( \mu^2 \):
$$ \mu^2 = \frac{5}{4} $$
Take the square root of both sides:
$$ \mu = \frac{\sqrt{5}}{2} $$
Conclusion
The refractive index of the prism is determined to be \(\frac{\sqrt{5}}{2}\).