Question

A parallel beam shifts laterally by 3.14 cm when it is incident on 90 cm thick slab at incident angle of 3º then refraction index of slab (nearest integer) is __________.

Answer 1

3

Answer 2

The lateral displacement ($\Delta$) of a light ray passing through a parallel slab of thickness ($t$) and refractive index ($\mu$) at a small angle of incidence ($\theta$) is given by the formula: $\Delta = \frac{t\theta(\mu - 1)}{\mu}$ Convert the angle of incidence from degrees to radians: $\theta = 3^\circ = 3 \times \frac{\pi}{180} \text{ radians} = \frac{\pi}{60} \text{ radians}$ Given $\Delta = 3.14$ cm, $t = 90$ cm, and using $\pi \approx 3.14$, we have $\theta \approx \frac{3.14}{60}$ radians. Substituting these values: $3.14 = \frac{90 \times \left(\frac{3.14}{60}\right) \times (\mu - 1)}{\mu}$ $1 = 1.5 \times \frac{\mu - 1}{\mu}$ $\mu = 1.5(\mu - 1) \implies \mu = 1.5\mu - 1.5 \implies 0.5\mu = 1.5 \implies \mu = 3$