a
As beam of light is incident normally on the face \(AB\) of the right angled prism \(ABC,\) so no refraction occurs at face \(AB\) and it passes straight and strikes the face \(AC\) at an angle of incidence, \(i=45^o\)

For total reflection to take place at face \(AC\),

\(i > i_{c}\)  or  \(\sin i>\sin i_{c}\)

where \(i_{c}\) is the critical angle. 

But as here \(i=45^o\) and \(\sin i_{c}=\frac{1}{\mu}\)

\(\therefore \quad \sin 45^o > \frac{1}{\mu}\) or \(\frac{1}{\sqrt{2}} > \frac{1}{\mu} \quad\) or \(\quad \mu>\sqrt{2}=1.414\)

As \(\mu_{\text {red }}(=1.39) < \mu(=1.414)\) while \(\mu_{\text {green }}(=1.44)\) and \(\mu_{\text {blue }}(=1.47) > \mu(=1.414),\) so only red colour will be transmitted through face \(A C\) while green and blue colours will suffer total internal reflection.

So the prism will separate red colour from the green and blue colours as shown in the following figure.