A ray of light is incident at an angle \( i \) on a parallel sided glass slab of thickness \( d \) and gets refracted into the slab at angle \( r \). Draw a ray diagram to show its path as it emerges out of the slab. Hence, obtain an expression for the lateral shift of the ray. Under what condition will the shift be minimum?
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The lateral shift of the ray inside a parallel sided slab depends on the thickness of the slab, the angle of incidence, and the angle of refraction. The shift will be minimum when the incident ray is normal to the slab surface (i.e., \( i = r \)).
Consider the ray diagram for a parallel-sided glass slab:
1. A light ray strikes the glass slab at an angle of incidence \( i \).
2. Inside the glass, the ray refracts at an angle \( r \).
3. The ray exits the slab at the same angle \( r \) due to its parallel sides.
The diagram shows that upon exiting the slab, the light ray is laterally shifted by a distance due to the slab's thickness \( d \).
Derivation of Lateral Shift:
Let \( x \) denote the lateral shift, which is the horizontal displacement of the emergent ray from its original path.
The light undergoes two refractions:
1. Upon entering the slab at angle \( i \), it refracts to angle \( r \).
2. Upon exiting the slab, it refracts back to angle \( r \).
Given the slab thickness \( d \), the distance traveled within the slab is \( d \sec r \) (as the ray travels at angle \( r \)).
The formula for the lateral shift \( x \) is:
\[ x = d \left( \frac{\sin(i - r)}{\cos r} \right) \]
where:
- \( i \) is the angle of incidence.
- \( r \) is the angle of refraction.
- \( d \) is the thickness of the slab.
Condition for Minimum Shift:
The lateral shift is minimized when the angle of incidence \( i \) equals the angle of refraction \( r \), i.e., \( i = r \). In this scenario, the ray passes straight through the slab without any lateral shift.
Therefore, the minimum lateral shift occurs when the incident angle equals the refracted angle (\( i = r \)).
