Question

It was found that the refractive index of material of a certain prism varied as $1.5+0.004/{\lambda }^2$, where $\lambda$ is the wavelength of light used to measure the refractive index. The same material was then used to construct a thin prism of apex angle ${10}^o$. Angles of minimum deviation $\left({\delta }_m\right)$ of the prism were recorded for the sources with wavelength ${\lambda }_1$ and ${\lambda }_2$ respectively. Then.

• A. ${\delta }_m\left({\lambda }_1\right)<{\delta }_m\left({\lambda }_2\right)$ if ${\lambda }_1<{\lambda }_2$
• B. ${\delta }_m\left({\lambda }_1\right)>{\delta }_m\left({\lambda }_2\right)$ if ${\lambda }_1>{\lambda }_2$
• C. ${\delta }_m\left({\lambda }_1\right)>{\delta }_m\left({\lambda }_2\right)$ if ${\lambda }_1<{\lambda }_2$
• D. ${\delta }_m$ is the same in both the cases

Answer 1

The correct option is C ${\delta }_m\left({\lambda }_1\right)>{\delta }_m\left({\lambda }_2\right)$ if ${\lambda }_1<{\lambda }_2$

Angle of minimum deviation ${\delta }_m=(\mu -1)A$
So, more is the refractive index of the material, lesser is the deviation.
Also, more is the wavelength of light, lesser is the refractive inde of material.
$⟹{\delta }_m\left({\lambda }_1\right)>{\delta }_m\left({\lambda }_2\right)$ if ${\lambda }_1<{\lambda }_2$