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A ray of light is incident on a parallel...

A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence `theta` is small, than the lateral displacement in the incident and emergent ray will be

A

`(t theta(n-1))/(n)`

B

`(t theta)/(n)`

C

`(t theta n)/(n-1)`

D

`(t theta(n+1))/(n)`

Text Solution

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The correct Answer is:
To find the lateral displacement of a ray of light passing through a parallel slab of thickness \( t \) and refractive index \( n \) when the angle of incidence \( \theta \) is small, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Geometry**: - Consider a parallel slab of thickness \( t \) and refractive index \( n \). - A ray of light strikes the slab at an angle \( \theta \) (the angle of incidence). 2. **Refraction at the Slab**: - When the ray enters the slab, it bends towards the normal due to the higher refractive index of the slab compared to air. - Let \( R \) be the angle of refraction in the slab. 3. **Using Snell's Law**: - According to Snell's Law: \[ n_1 \sin \theta = n_2 \sin R \] - Here, \( n_1 = 1 \) (refractive index of air) and \( n_2 = n \) (refractive index of the slab). - Thus, we have: \[ \sin \theta = n \sin R \] - For small angles, we can approximate \( \sin \theta \approx \theta \) and \( \sin R \approx R \). 4. **Finding \( R \)**: - From the equation \( \sin \theta = n \sin R \), we can express \( \sin R \): \[ \sin R = \frac{\sin \theta}{n} \approx \frac{\theta}{n} \] 5. **Calculating Lateral Displacement**: - The lateral displacement \( D \) can be calculated using the formula: \[ D = \frac{t}{\cos R} \left( \sin \theta - \sin R \right) \] - For small angles, \( \cos R \approx 1 \) and \( \sin R \approx \frac{\theta}{n} \): \[ D \approx t \left( \theta - \frac{\theta}{n} \right) \] - Simplifying this gives: \[ D \approx t \theta \left( 1 - \frac{1}{n} \right) \] 6. **Final Expression for Lateral Displacement**: - Therefore, the lateral displacement \( D \) is: \[ D = \frac{t \theta (n - 1)}{n} \] ### Conclusion: The lateral displacement of the incident and emergent ray is given by: \[ D = \frac{t \theta (n - 1)}{n} \]
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