With respect to air,the critical angle in a m | vedclass

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(C) The critical angle $C$ is given by the formula $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the medium.Since the refractive index $

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less than $	heta$

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(C) The critical angle $C$ is given by the formula $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the medium.Since the refractive index $n$ is inversely proportional to the wavelength $\lambda$ (Cauchy's relation: $n \approx A + \frac{B}{\lambda^2}$),as the wavelength decreases,the refractive index increases.For red light,the wavelength is $\lambda_1$. For yellow light,the wavelength is $\lambda_2$. Since $\lambda_1 > \lambda_2$,the refractive index for red light $(n_r)$ is less than the refractive index for yellow light $(n_y)$.Therefore,$n_r Since $\sin C = \frac{1}{n}$,a higher refractive index results in a smaller critical angle.Thus,the critical angle for yellow light will be less than the critical angle for red light $(	heta)$.

With respect to air,the critical angle in a medium for red light of wavelength $\lambda_1$ is $\theta$. Other facts remaining same,the critical angle for yellow light of wavelength $\lambda_2$ will be

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