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A ray of light is incident at an angle of 30o on one top surface...

Physics
JAMB 1987

A ray of light is incident at an angle of 30o on one top surface of a parallel-sided glass block of refractive index 1.5. The ray finally emerges from the lower surface. What is the angular deviation of the emergent ray?

  • A. 60o
  • B. 39o
  • C. 28o
  • D. 0o
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Correct Answer: Option D
Explanation

To find the angular deviation of a ray of light passing through a parallel-sided glass block, follow these steps:

Calculate the angle of refraction at the first surface: Using Snell's Law:
\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)
where:
\(n_1 = 1 \quad (\text{air}), \quad \theta_1 = 30^\circ, \quad n_2 = 1.5 \quad (\text{glass})\)
Substituting the values:
\(1 \cdot \sin(30^\circ) = 1.5 \cdot \sin(\theta_2) \Rightarrow 0.5 = 1.5 \cdot \sin(\theta_2) \Rightarrow \sin(\theta_2) = \frac{1}{3}\)
Thus,
\(\theta_2 \approx \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\)

Angle of incidence at the second surface: Since the block is parallel-sided, the angle of incidence is: \(\theta_3 = \theta_2 \approx 19.47^\circ\)

Calculate the angle of refraction at the second surface: Applying Snell's Law again: \(n_2 \sin(\theta_3) = n_1 \sin(\theta_4)\)
Substituting values: \(1.5 \cdot \sin(19.47^\circ) = 1 \cdot \sin(\theta_4) \Rightarrow \sin(\theta_4) \approx 0.5 \Rightarrow \theta_4 = \arcsin(0.5) = 30^\circ\)

Calculate the angular deviation: The angular deviation \(D\) is given by: \(D = \theta_1 - \theta_4 = 30^\circ - 30^\circ = 0^\circ\)

The angular deviation of the emergent ray is \(0^\circ\).


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