A ray of monochromatic light in air falls on the upper surface of two parallel-sided slabs of transparent material, as shown in the diagram. What is the angle of refraction f in the lower slab? 60° n1- 2.30 n2-1.73
The Correct Answer and Explanation is :
To find the angle of refraction ((f)) in the lower slab, we need to use Snell’s Law, which is expressed as:
[
n_1 \sin \theta_1 = n_2 \sin \theta_2
]
Where:
- (n_1) and (n_2) are the refractive indices of the materials,
- (\theta_1) and (\theta_2) are the angles of incidence and refraction, respectively.
Step 1: Refraction at the air-upper slab interface
Given:
- Angle of incidence ((\theta_1)) = 60°,
- Refractive index of air ((n_{\text{air}})) = 1,
- Refractive index of the upper slab ((n_1)) = 2.30.
Using Snell’s Law:
[
n_{\text{air}} \sin 60^\circ = n_1 \sin \theta_2
]
[
1 \cdot \sin 60^\circ = 2.30 \cdot \sin \theta_2
]
[
\sin \theta_2 = \frac{\sin 60^\circ}{2.30}
]
[
\sin \theta_2 = \frac{0.866}{2.30} = 0.376
]
[
\theta_2 = \arcsin(0.376) \approx 22.1^\circ
]
Step 2: Refraction at the upper-lower slab interface
The light passes into the second slab. Here:
- Angle of incidence ((\theta_2′)) = (22.1^\circ),
- Refractive index of upper slab ((n_1)) = 2.30,
- Refractive index of lower slab ((n_2)) = 1.73.
Using Snell’s Law:
[
n_1 \sin \theta_2′ = n_2 \sin f
]
[
2.30 \cdot \sin 22.1^\circ = 1.73 \cdot \sin f
]
[
\sin f = \frac{2.30 \cdot \sin 22.1^\circ}{1.73}
]
[
\sin f = \frac{2.30 \cdot 0.376}{1.73} = \frac{0.865}{1.73} \approx 0.500
]
[
f = \arcsin(0.500) = 30^\circ
]
Final Answer:
The angle of refraction ((f)) in the lower slab is 30°.
Explanation:
- Snell’s Law governs how light bends when moving between materials with different refractive indices.
- In the first interface, light bends toward the normal because it moves into a denser medium ((n_1 = 2.30)).
- At the second interface, the light bends away from the normal because it enters a less dense medium ((n_2 = 1.73)).
- By systematically applying Snell’s Law at each boundary, the final angle of refraction ((f)) is found to be 30°.