Question

White light is incident normally on a glass plate of thickness 0.50 × $10^{-6}$ and index of refraction 1.50. Which wavelengths in the visible region (400 nm-700 nm) are strongly reflected by the plate?

Answer 1

500 nm

Answer 2

The phenomenon is thin-film interference. For normal incidence, the optical path difference (OPD) between the two reflected rays is $2\mu t$.

Given: Thickness of the plate, $t = 0.50 \times 10^{-6}$ m Refractive index of the plate, $\mu = 1.50$ Visible light range: $\lambda \in [400 \, \text{nm}, 700 \, \text{nm}]$.

The optical path difference is: $2\mu t = 2 \times 1.50 \times (0.50 \times 10^{-6} \, \text{m}) = 1.5 \times 10^{-6} \, \text{m} = 1500 \, \text{nm}$.

Phase Shifts:

1. Reflection at the first surface (air to glass): $\mu_{glass} > \mu_{air}$, so there is a phase change of $\pi$.
2. Reflection at the second surface (glass to air): $\mu_{glass} > \mu_{air}$, so there is a phase change of $\pi$.

When both reflections involve a phase shift of $\pi$, the condition for constructive interference (strong reflection) is $2\mu t = n\lambda$, where $n$ is an integer.

We need to find integer values of $n$ for which $\lambda$ falls within the visible region (400 nm to 700 nm): $\lambda = \frac{2\mu t}{n} = \frac{1500 \, \text{nm}}{n}$.

$400 \le \frac{1500}{n} \le 700$

From $400 \le \frac{1500}{n}$: $n \le \frac{1500}{400} = 3.75$. From $\frac{1500}{n} \le 700$: $n \ge \frac{1500}{700} \approx 2.14$.

The possible integer value for $n$ is $3$. For $n=3$, $\lambda = \frac{1500}{3} = 500 \, \text{nm}$.

Therefore, the wavelength strongly reflected is 500 nm.