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Light of wavelength $5000 Å$ falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Solution:
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Verified Answer
Wavelength of incident light $=5000 Å$.
There is no change in the wavelength and the frequency of reflected light.
$\therefore \quad \lambda=5000 Å$, frequency,
$$
v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{5000 \times 10^{-10}}=6 \times 10^{14} \mathrm{~Hz}
$$
When the reflected ray is normal to the incident ray,
$$
\begin{aligned}
& \angle \mathrm{i}+\angle \mathrm{r}=90^{\circ} \\
\because \quad \angle \mathrm{i}=\angle \mathrm{r} \\
\therefore \quad \angle \mathrm{i}+\angle \mathrm{i}=90^{\circ} \Rightarrow 2 \mathrm{i}=90^{\circ} \quad \therefore \quad \mathrm{i}=45^{\circ}
\end{aligned}
$$
There is no change in the wavelength and the frequency of reflected light.
$\therefore \quad \lambda=5000 Å$, frequency,
$$
v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{5000 \times 10^{-10}}=6 \times 10^{14} \mathrm{~Hz}
$$
When the reflected ray is normal to the incident ray,
$$
\begin{aligned}
& \angle \mathrm{i}+\angle \mathrm{r}=90^{\circ} \\
\because \quad \angle \mathrm{i}=\angle \mathrm{r} \\
\therefore \quad \angle \mathrm{i}+\angle \mathrm{i}=90^{\circ} \Rightarrow 2 \mathrm{i}=90^{\circ} \quad \therefore \quad \mathrm{i}=45^{\circ}
\end{aligned}
$$
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