Question:

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

Updated On: Mar 26, 2025
  • \(\frac{\sqrt5}{4}\)
  • \(\frac{\sqrt5}{2}\)
  • \(\frac{\sqrt3}{4}\)
  • \(\frac{\sqrt3}{2}\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

 

Step 1: Relation in Prism

From the prism relation:

$$ r_1 + c = A $$

Rearranging for \( r_1 \):

$$ r_1 = 90^\circ - c \quad \text{...(1)} $$

Step 2: Expression for \( \cos c \)

We know:

$$ \sin c = \frac{1}{\mu} $$

Using trigonometric identity:

$$ \cos c = \frac{\sqrt{\mu^2 - 1}}{\mu} $$

Step 3: Apply Snell's Law on Incidence Surface

Using Snell’s law at the first surface:

$$ \sin 30^\circ = \mu \sin (r_1) $$

Substituting \( r_1 = 90^\circ - c \):

$$ \frac{1}{2} = \mu \sin (90^\circ - c) $$

Since \( \sin (90^\circ - c) = \cos c \), we get:

$$ \frac{1}{2} = \mu \times \frac{\sqrt{\mu^2 - 1}}{\mu} $$

Simplifying:

$$ \frac{1}{2} = \frac{\sqrt{\mu^2 - 1}}{1} $$

Step 4: Solve for \( \mu \)

Squaring both sides:

$$ \frac{1}{4} = \mu^2 - 1 $$

Rearranging:

$$ \mu^2 = \frac{5}{4} $$

Taking square root:

$$ \mu = \frac{\sqrt{5}}{2} $$

Conclusion

The refractive index \( \mu \) of the prism is \(\frac{\sqrt{5}}{2}\).

Was this answer helpful?
4
28