The following figure represents a wave front | vedclass

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(A) In the given figure,$AB$ is the incident wave front and $CD$ is the refracted wave front.Let $i$ be the angle of incidence and $r$ be the angle of

Vedclass · Accepted Answer

$\frac{\cos 	heta_1}{\cos 	heta_4}$

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(A) In the given figure,$AB$ is the incident wave front and $CD$ is the refracted wave front.Let $i$ be the angle of incidence and $r$ be the angle of refraction.The angle between the incident wave front $AB$ and the boundary $PQ$ is $	heta_1$. Since the ray is perpendicular to the wave front,the angle of incidence $i = 90^\circ - 	heta_1$.Similarly,the angle between the refracted wave front $CD$ and the boundary $PQ$ is $	heta_4$. The angle of refraction $r = 90^\circ - 	heta_4$.According to Snell's law,the refractive index of the medium $\mu$ with respect to air is given by $\mu = \frac{\sin i}{\sin r}$.Substituting the values of $i$ and $r$:$\mu = \frac{\sin(90^\circ - 	heta_1)}{\sin(90^\circ - 	heta_4)}$$\mu = \frac{\cos 	heta_1}{\cos 	heta_4}$

The following figure represents a wave front $AB$ which passes from air to another transparent medium and produces a new wave front $CD$ after refraction. The refractive index of the medium is $:$ ($PQ$ is the boundary between air and the medium)

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