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Light of wavelength 5000 Armstrong falls

Question:

Light of wavelength 5000 Armstrong 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:

Wavelength of incident light, $[\lambda]=5000$ Armstrong $=5000 \times 10^{-10} \mathrm{~m}$

Speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m}$

Following is the relation for the frequency of incident light:

$\mathrm{v}=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{5000 \times 10^{-10}}=6 \times 10^{14}$

The wavelength and frequency of incident light is equal to the reflected ray. Therefore, 5000 Armstrong and

$6 \times 10^{14} \mathrm{~Hz}$ is the wavelength and frequency of the reflected light. When reflected ray is normal to incident

ray, the sum of the angle of incidence, $\angle i$ and angle of reflection, $\angle r$ is $90^{\circ}$.

From laws of reflection we know that the angle of incidence is always equal to the angle of reflection

$\angle i+\angle r=90^{\circ}$

i.e. $\angle i+\angle i=90^{\circ}$

Therefore, 45° is the angle of incidence.

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