A monochromatic light wave with wavelength la | vedclass

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(B) The frequency of a light wave depends only on the source and remains constant when it travels from one medium to another. Therefore,$v_2 = v_1$.Ac

Vedclass · Accepted Answer

$\lambda_2=\frac{1}{\sqrt{2}} \lambda_1, v_2=v_1$

Vedclass · Answer

(B) The frequency of a light wave depends only on the source and remains constant when it travels from one medium to another. Therefore,$v_2 = v_1$.According to Snell's law,$\mu_1 \sin i = \mu_2 \sin r$,where $\mu_1$ and $\mu_2$ are the refractive indices of air and the medium respectively.Given $i = 45^{\circ}$ and $r = 30^{\circ}$,and knowing $\mu_1 \approx 1$ for air:$1 \cdot \sin 45^{\circ} = \mu_2 \cdot \sin 30^{\circ}$$\frac{1}{\sqrt{2}} = \mu_2 \cdot \frac{1}{2}$$\mu_2 = \sqrt{2}$.Since the refractive index $\mu = \frac{c}{v} = \frac{\lambda_1 f}{\lambda_2 f} = \frac{\lambda_1}{\lambda_2}$,we have $\mu_2 = \frac{\lambda_1}{\lambda_2}$.Substituting $\mu_2 = \sqrt{2}$,we get $\sqrt{2} = \frac{\lambda_1}{\lambda_2}$,which implies $\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$.

$A$ monochromatic light wave with wavelength $\lambda_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $45^{\circ}$ and $30^{\circ}$ respectively,then the wavelength $\lambda_2$ and frequency $v_2$ of the refracted wave are :

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