A right-angled prism \( ABC \) (refractive index \( \sqrt{2} \)) is kept on a plane mirror as shown in the figure. A ray of light is incident normally on the face \( AC \). Find the angle of deviation produced by the prism.
Show Hint
Check TIR at AB (60 degrees is above the 45 degree critical angle), then the mirror, then AC again at normal incidence. The ray comes back along the incident direction.
Step 1: Understanding the Concept:
The angle of deviation is the angle between the direction of the ray that enters the prism and the direction of the ray that finally leaves it.
So we follow the ray through every surface it meets. From the figure, \(\angle A = 60^{\circ}\) and \(\angle B = 90^{\circ}\), so \(\angle C = 30^{\circ}\).
Step 2: Key Formula or Approach:
A ray goes through total internal reflection (TIR) at a glass-air surface when its angle of incidence is more than the critical angle \(i_c\).
\[ \sin i_c = \frac{1}{n} = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad i_c = 45^{\circ} \]
Step 3: Face AC:
The ray falls normally on AC, so the angle of incidence is \(0^{\circ}\). It goes into the prism without bending.
Step 4: Face AB:
The ray travels along the normal to AC. The normal to AC and the normal to AB make the same angle as the faces AC and AB make with each other, which is \(\angle A = 60^{\circ}\).
So the angle of incidence on AB is \(60^{\circ}\). This is more than \(i_c = 45^{\circ}\), so the ray is totally reflected at AB.
After reflection the ray makes \(60^{\circ}\) with the normal of AB, which means \(30^{\circ}\) with the face AB itself.
Step 5: Plane mirror along BC:
AB is perpendicular to BC, so the normal to BC is parallel to AB. The ray makes \(30^{\circ}\) with AB, so it also makes \(30^{\circ}\) with the normal of BC.
The angle of incidence on the mirror is \(30^{\circ}\), and the ray is reflected at \(30^{\circ}\) on the other side of the normal. It now moves up towards AC.
Step 6: Back at face AC:
Look at the triangle made by the reflected ray, the face BC and the face AC. The ray makes \(60^{\circ}\) with BC, and \(\angle C = 30^{\circ}\).
The third angle is \(180^{\circ} - 60^{\circ} - 30^{\circ} = 90^{\circ}\). So the ray hits AC at right angles, with zero angle of incidence, and leaves the prism without bending.
Path in words: enters AC straight, is totally reflected at AB, is reflected by the mirror BC, and comes out of AC along the normal, exactly opposite to the incident ray.
Final Answer:
The emergent ray is parallel and opposite to the incident ray, so the angle between the two directions is \(180^{\circ}\).
\[ \boxed{\delta = 180^{\circ}} \]
