JAMB - Physics (1987 - No. 30)

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?
60o
39o
28o
0o

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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