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

The wavelengths in the visible region (400 nm-700 nm) that are strongly reflected by the plate are approximately 428.57 nm and 600 nm.

Answer 2

For normal incidence, the optical path difference (OPD) between the two reflected rays is $2\mu t$. There is a phase shift of $\pi$ at the first air-glass interface (since $\mu_{glass} > \mu_{air}$) and no phase shift at the second glass-air interface (since $\mu_{air} < \mu_{glass}$). The net phase difference due to reflections is $\pi$.

For constructive interference (strong reflection), the condition is: $2\mu t = (m + 1/2)\lambda$

Given: $t = 0.50 \times 10^{-6}$ m $\mu = 1.50$

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

The condition becomes: $1500 \, \text{nm} = (m + 0.5)\lambda$ $\lambda = \frac{1500}{m + 0.5} \, \text{nm}$

We need to find wavelengths $\lambda$ in the visible region (400 nm - 700 nm). For $\lambda_{min} = 400$ nm: $400 = \frac{1500}{m + 0.5} \implies m + 0.5 = \frac{1500}{400} = 3.75 \implies m = 3.25$.

For $\lambda_{max} = 700$ nm: $700 = \frac{1500}{m + 0.5} \implies m + 0.5 = \frac{1500}{700} \approx 2.143 \implies m \approx 1.643$.

The integer values for $m$ in the range $[1.643, 3.25]$ are $m=2$ and $m=3$.

For $m=2$: $\lambda = \frac{1500}{2 + 0.5} = \frac{1500}{2.5} = 600 \, \text{nm}$.

For $m=3$: $\lambda = \frac{1500}{3 + 0.5} = \frac{1500}{3.5} = \frac{3000}{7} \approx 428.57 \, \text{nm}$.

These are the wavelengths strongly reflected.