When light is refracted into a denser medium,

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(C) According to the formula of refractive index,$_1\mu_2 = \dfrac{v_1}{v_2} = \dfrac{v_{	ext{rarer}}}{v_{	ext{denser}}} \Rightarrow v_1 > v_2$. The

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its wavelength decreases but frequency remains unchanged.

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(C) According to the formula of refractive index,$_1\mu_2 = \dfrac{v_1}{v_2} = \dfrac{v_{	ext{rarer}}}{v_{	ext{denser}}} \Rightarrow v_1 > v_2$. The speed of light is given by $v = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength. When light travels from one medium to another,its frequency $f$ remains unchanged because it depends on the source of light. Therefore,$f_1 = f_2 = f$. Since the speed $v_1 > v_2$,it follows that $\lambda_1 f > \lambda_2 f$,which implies $\lambda_1 > \lambda_2$. Thus,when light enters a denser medium,its wavelength decreases while the frequency remains constant. Therefore,option $C$ is correct.

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