1. Use the condition of no reflection for blue light

Since there is no reflection at the first surface, the incidence is at Brewster angle.

For air to prism transition, $an i_B = \mu_b$ Given for blue light, $\mu_b = \sqrt{3}$ So, $i_B = 60^\circ$

At Brewster angle, the reflected and refracted rays are perpendicular, hence $i_B + r_1 = 90^\circ$ Therefore, $r_1 = 30^\circ$

2. Use deviation formula for blue light

For a prism, $\delta = i + e - A$ where:

• $i$ = angle of incidence,
• $e$ = angle of emergence,
• $A$ = prism angle.

Also inside the prism, $r_1 + r_2 = A$ So here, $r_2 = A - 30^\circ$

Given blue-light deviation, $\delta = 60^\circ$ Hence, $60^\circ = 60^\circ + e - A$ which gives $e = A$

Now apply Snell's law at the second surface: $\mu_b \sin r_2 = \sin e$ $\sqrt{3}\sin(A-30^\circ)=\sin A$

Expand: $\sqrt{3}\left(\sin A\cos30^\circ - \cos A\sin30^\circ\right)=\sin A$ $\sqrt{3}\left(\sin A\cdot \frac{\sqrt{3}}{2}-\cos A\cdot \frac12\right)=\sin A$ $\frac{3}{2}\sin A - \frac{\sqrt{3}}{2}\cos A = \sin A$ $\frac12 \sin A = \frac{\sqrt{3}}{2}\cos A$ $\tan A = \sqrt{3}$ Thus, $A=60^\circ$

So option B ($A=45^\circ$) is false.

Also from above, $e=A=60^\circ$ Hence the angle of refraction (emergence into air) at the exit face is $60^\circ$. So D is correct.

3. Check polarization direction

At Brewster angle, reflected light is polarized perpendicular to plane of incidence, and the transmitted/refracted light is therefore polarized in the plane of incidence.

Since the incident blue light is plane polarized and arranged so that there is no reflection, its electric vector must be in the plane of incidence.

So A is correct.

4. Use minimum deviation for red light

Given for red light, $\delta_{\min}=30^\circ$ For a prism at minimum deviation, $\mu_r = \frac{\sin\left(\frac{A+\delta_{\min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Using $A=60^\circ$,