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(C) For a thin prism,the angle of minimum deviation $\delta_{m}$ is given by the formula: $\delta_{m} = (\mu - 1)A$,where $A$ is the apex angle.Given

Vedclass · Accepted Answer

$\delta_{m}(\lambda_{1}) > \delta_{m}(\lambda_{2})$ if $\lambda_{1} < \lambda_{2}$

Vedclass · Answer

(C) For a thin prism,the angle of minimum deviation $\delta_{m}$ is given by the formula: $\delta_{m} = (\mu - 1)A$,where $A$ is the apex angle.Given $\mu = 1.5 + \frac{0.004}{\lambda^{2}}$,we substitute this into the formula:$\delta_{m} = \left(1.5 + \frac{0.004}{\lambda^{2}} - 1\right)A = \left(0.5 + \frac{0.004}{\lambda^{2}}\right)A$.From this expression,it is clear that $\delta_{m}$ is inversely proportional to $\lambda^{2}$.Therefore,as the wavelength $\lambda$ increases,the refractive index $\mu$ decreases,which in turn causes the angle of minimum deviation $\delta_{m}$ to decrease.If $\lambda_{1}  \mu(\lambda_{2})$.Consequently,$\delta_{m}(\lambda_{1}) > \delta_{m}(\lambda_{2})$.

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