The wavelength of light in two liquids x and | vedclass

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(C) The refractive index $n$ of a medium is inversely proportional to the wavelength $\lambda$ of light in that medium,given by $n = \frac{c}{v} = \fr

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) The refractive index $n$ of a medium is inversely proportional to the wavelength $\lambda$ of light in that medium,given by $n = \frac{c}{v} = \frac{\lambda_0}{\lambda}$.For two media $x$ and $y$,the relative refractive index of $x$ with respect to $y$ is given by $n_{xy} = \frac{n_x}{n_y} = \frac{\lambda_y}{\lambda_x}$.Given $\lambda_x = 3500 \ \mathring{A}$ and $\lambda_y = 7000 \ \mathring{A}$.Since light travels from a denser medium to a rarer medium for total internal reflection to occur,the denser medium is $x$ (smaller wavelength) and the rarer medium is $y$ (larger wavelength).The critical angle $C$ is defined by $\sin C = \frac{n_y}{n_x} = \frac{\lambda_x}{\lambda_y}$.Substituting the values: $\sin C = \frac{3500}{7000} = \frac{1}{2}$.Therefore,$C = \arcsin(0.5) = 30^o$.

The wavelength of light in two liquids $x$ and $y$ is $3500 \ \mathring{A}$ and $7000 \ \mathring{A}$ respectively. The critical angle of $x$ relative to $y$ will be.......$^o$

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