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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 :
 

Option: 1

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


Option: 2

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


Option: 3

\lambda_2=\lambda_1, v_2=\sqrt{2} v_1


Option: 4

\lambda_2=\sqrt{2} \lambda_1, v_2=v_1


Answers (1)

best_answer

\begin{aligned} & 1 \times \sin 45=\mu \sin 30 \\ & \Rightarrow \frac{1}{\sqrt{2}}=\mu \times \frac{1}{2} \\ & \Rightarrow \mu=\sqrt{2}---(i) \\ & \text { Now, } \frac{\mu_1}{\mu_2}=\frac{V_2}{V_1}=\frac{\lambda_2}{\lambda_1}----(i i) \\ & \text { Using \ \ eq (i) and (ii), }\\ &Using eq (i) \ \ and \ \ (ii),\\ &\lambda_2=\frac{1}{\sqrt{2}} \lambda_1\\ & \ And \ \ V_2=\frac{1}{\sqrt{2}} V_1\\ &\text{ Now, for relation between frequencies,}\\ &Frequency, v=\frac{V}{\lambda}\\ & \text { Or } \frac{v_1}{v_2}=\frac{v_1}{v_2} \times \frac{\lambda_2}{\lambda_1}=1 \\ & v_1=v_2 \end{aligned}

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