Question

A ray of light is incident on a parallel slab of thickness t and refractive index $\mu$. If the angle of incidence is $\theta$ then for small $\theta$ show that the lateral displacement of light ray will be $\Delta = \frac{t\theta(\mu - 1)}{\mu}$.

Answer 1

The lateral displacement of the light ray for small angle of incidence $\theta$ is $\Delta = \frac{t\theta(\mu - 1)}{\mu}$.

Answer 2

1. Snell's Law: $1 \cdot \sin \theta = \mu \sin \phi$.
2. Lateral Displacement Formula: $\Delta = \frac{t \sin(\theta - \phi)}{\cos \phi}$.
3. Small Angle Approximation: For small $\theta$, $\sin \theta \approx \theta$, $\cos \phi \approx 1$, and $\phi \approx \theta/\mu$ (from Snell's Law: $\theta \approx \mu \phi$).
4. Substitution: $\Delta \approx t(\theta - \phi) \approx t(\theta - \theta/\mu) = \frac{t\theta(\mu-1)}{\mu}$.