In the diagram below, AB, BC, and BD represent the incident, reflected and refracted rays, respectively. When a plane wave front with wavelength 6000 Å is incident at the point of separation of air and denser medium, the wavelength of refracted light is _____.

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In the diagram below, AB, BC, and BD represent the incident, reflected and refracted rays, respectively. When a plane wave front with wavelength 6000 Å is incident at the point of separation of air and denser medium, the wavelength of refracted light is _____.

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3000 Å
3564 Å
4243 Å
5196 Å

Answer 1

Concept:

According to Snell law of refraction, the incident ray, refracted ray and the normal to the interface of two media at the point of incidence all lie in the same plane.

The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant which is equal to refractive index.

\(\mu=\frac{sini}{sinr}=\frac{\lambda_1}{\lambda_2}\)

Angle of incidence=45º
Angle of refraction=30º.

Explanation:

\(\frac{sini}{sinr}=\frac{\lambda_1}{\lambda_2}=\frac{sin45^o}{sin30^o}=\frac{6000}{\lambda_2}\)

\(\frac{sin45^o}{sin30^o}=\frac{6000}{\lambda_2}\)

Now, \(sin45^o=\frac{1}{\sqrt{2}}\text{ and } sin30^o=\frac{1}{2} \)

Substitute all values, we get,

\(\lambda_2=\frac{6000\sqrt{2}}{2}A^o=4243A^o\)

Hence, the correct answer is Option-3-\(\lambda_2=4243A^o\).

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In the diagram below, AB, BC, and BD represent the incident, reflected and refracted rays, respectively. When a plane wave front with wavelength 6000 Å is incident at the point of separation of air and denser medium, the wavelength of refracted light is _____.

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