Red light of wavelength 632 nm in free space is incident at an angle of $\t$theta = 39^\circ$ with respect to the normal on a glass slide of thickness d = 0.78 mm and index of refraction n = 1.52. The wavelength of light in glass will become

Correct Answer: (C) 416 nm

The wavelength of light in a medium with refractive index $n$ is given by ${\lambda }_{\text{medium}}=\frac{{\lambda }_0}{n}$$\lambda$_{$\text{medium}$} = $\frac{\lambda_0}{n}$. Substituting ${\lambda }_0=632\text{ nm}$$\lambda$_0 = 632 $\text{ nm}$ and $n = 1.52$, we get ${\lambda }_{\text{medium}}=\frac{632}{1.52}\approx 415.79\text{ nm}\approx 416\text{ nm}$$\lambda$_{$\text{medium}$} = $\frac{632}{1.52}$ $\approx$ 415.79 $\text{ nm}$ $\approx$ 416 $\text{ nm}$.