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. The angle of refraction f in the lower slab is approximately
The Correct Answer and Explanation is :
When a ray of monochromatic light in air strikes the upper surface of two parallel-sided slabs of transparent material, it undergoes refraction at each interface. The angle of refraction in the lower slab can be determined using Snell’s Law, which relates the angles of incidence and refraction to the refractive indices of the media involved.
Understanding Snell’s Law:
Snell’s Law states that:
[ \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_2}{n_1} ]
Where:
- ( \theta_1 ) is the angle of incidence in the first medium.
- ( \theta_2 ) is the angle of refraction in the second medium.
- ( n_1 ) and ( n_2 ) are the refractive indices of the first and second media, respectively.
Applying Snell’s Law to the Problem:
- First Refraction (Air to Upper Slab):
- Let the angle of incidence in air be ( \theta_1 ).
- The refractive index of air is approximately 1.
- The refractive index of the upper slab is ( n_1 ).
- Using Snell’s Law: [ \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_1}{1} ] Solving for ( \theta_2 ): [ \theta_2 = \sin^{-1}\left( \frac{\sin(\theta_1)}{n_1} \right) ]
- Second Refraction (Upper Slab to Lower Slab):
- The angle of incidence in the upper slab is ( \theta_2 ).
- The refractive index of the upper slab is ( n_1 ).
- The refractive index of the lower slab is ( n_2 ).
- Using Snell’s Law: [ \frac{\sin(\theta_2)}{\sin(\theta_3)} = \frac{n_2}{n_1} ] Solving for ( \theta_3 ): [ \theta_3 = \sin^{-1}\left( \frac{n_1}{n_2} \sin(\theta_2) \right) ]
- Final Refraction (Lower Slab to Air):
- The angle of incidence in the lower slab is ( \theta_3 ).
- The refractive index of the lower slab is ( n_2 ).
- The refractive index of air is 1.
- Using Snell’s Law: [ \frac{\sin(\theta_3)}{\sin(\theta_4)} = \frac{1}{n_2} ] Solving for ( \theta_4 ): [ \theta_4 = \sin^{-1}\left( \frac{n_2}{1} \sin(\theta_3) \right) ]
The angle of refraction in the lower slab, ( \theta_3 ), is determined by the refractive indices of the upper and lower slabs and the initial angle of incidence in air.
Conclusion:
The angle of refraction in the lower slab depends on the refractive indices of the materials and the initial angle of incidence. By applying Snell’s Law at each interface, one can calculate the angles of refraction at each stage of the light’s passage through the slabs.
For a visual explanation of refraction and Snell’s Law, you might find the following video helpful: