Answer to Question #227197 in Electricity and Magnetism for Arvin Shopon

Question

The refractive index of silicon depends on the wavelength of light according to the figure below. The light beam from the mercury discharge tube hits the silicon prism perpendicularly.

Light includes, among other things, a purple component (wavelength 405 nm) and a red component (wavelength 690 nm). The folding angle φ of the prism is 34.0°. What is the angle between the purple and red rays in the light passing through the prism?

Answer 1

Given,

Wavelength of purple light $(\lambda_p) =405nm$

Wavelength of red light component $(\lambda_r)=690nm$

Folding angle of prism $(\phi)=34.0^\circ$

As in the question there is some diagram, but it is not showing due to some error so i am taking some it $\mu$.

$\Rightarrow \theta_{p}=\sin^{-1}(\lambda_p \sin(\phi))$

$\Rightarrow \theta_{p}=\sin^{-1}(405\times 10^{-9}\sin(34))$

$\Rightarrow \theta_{p}=\sin^{-1}(405\times10^{-9}\times 0.56)$

$\Rightarrow \theta_{p}=\sin^{-1}(226.8\times 10^{-9})$

Similarly, for the wavelength of red light,

$\theta_{r}=\sin^{-1}(\lambda_r \sin(\phi))$

$\Rightarrow \theta_{r}=\sin^{-1}(690 \times 10^{-9} \sin(34))$

$\Rightarrow \theta_{r}=\sin^{-1}(690 \times 10^{-9} \times 0.56)$

$\Rightarrow \theta_{r}=\sin^{-1}(386.4 \times 10^{-9} )$

Hence, the resultant angle,

$\phi=\theta_p -\theta_r =\sin^{-1}(226.8\times 10^{-9}) - \sin^{-1}(386.4 \times 10^{-9} )$