A ray of light QP is incident normally on the face BC of a triangular prism ABC of refractive index 1.5 kept in air, as shown in the figure. Trace the path of the ray as it passes through the prism and give relevant explanation.
Show Hint
In prism tracing problems, always calculate the angles inside the small geometric triangles formed by the ray paths and the prism faces. Knowing that the angle between the ray and the face plus the angle of incidence equals 90 degrees is the ultimate key to solving these flawlessly.
Concept: • When a light ray strikes an optical boundary perfectly perpendicularly (normal incidence, $i = 0^\circ$), it proceeds entirely without deviation.
• At an internal boundary, if the angle of incidence strictly exceeds the critical angle ($i > C$), Total Internal Reflection (TIR) occurs rather than refraction.
• The critical angle formula for a glass-air interface is universally given by $\sin C = \frac{1}{\mu}$.
Step 1: Calculate the critical angle of the prism
The prism is composed of glass with a given refractive index of $\mu = 1.5$.
We first decisively calculate the critical angle ($C$) for the glass-air interface using the established formula:
\[ \sin C = \frac{1}{\mu} \]
\[ \sin C = \frac{1}{1.5} = \frac{2}{3} \approx 0.667 \]
Since $\sin 41.8^\circ \approx 0.667$, the critical angle is firmly established as $C \approx 41.8^\circ$.
Based on standard optical diagram conventions for such problems, we assume the prism ABC is an equilateral triangle, meaning all interior vertex angles are exactly $A = B = C = 60^\circ$. Step 2: Trace ray behavior at the first interface (BC)
The problem states that the incoming ray QP is incident exactly normally (perpendicularly) on the bottom face BC.
Because the angle of incidence is exactly $0^\circ$ relative to the surface normal, the ray enters the dense glass perfectly undeviated.
It travels vertically straight upward through the solid interior of the prism. Step 3: Trace ray behavior at the second interface (AB)
The vertically traveling ray then strikes the slanted interior face AB.
Using simple geometry inside the right-angled triangle formed by the vertical ray, the horizontal base BC, and the slanted face AB: the angle at vertex B is $60^\circ$, the angle between the ray and the base is $90^\circ$, therefore the angle between the vertical ray and the face AB is $180^\circ - 90^\circ - 60^\circ = 30^\circ$.
The surface normal to face AB is drawn perfectly perpendicular ($90^\circ$) to it.
Thus, the angle of incidence $i$ relative to this normal is mathematically $90^\circ - 30^\circ = 60^\circ$.
We carefully compare this angle of incidence to our previously calculated critical angle:
$i = 60^\circ$, while $C = 41.8^\circ$.
Since $i > C$, the ray is completely trapped and undergoes Total Internal Reflection (TIR) inside the prism.
According to the law of reflection, the ray reflects at an exact angle of $60^\circ$ from the normal, which simultaneously means it makes a geometric angle of $90^\circ - 60^\circ = 30^\circ$ with the face AB itself. Step 4: Trace ray behavior at the third interface (AC)
The newly reflected ray travels straight across the upper portion of the prism towards the opposite slanted face AC.
We now examine the small geometric triangle formed by the top vertex A, the point of TIR on face AB, and the new point of incidence on face AC.
Inside this specific triangle:
The top vertex angle $A$ is $60^\circ$.
The angle between the reflected ray and face AB was just established as $30^\circ$.
The sum of angles in any triangle must be $180^\circ$. Therefore, the third angle (where the ray strikes face AC) must strictly be $180^\circ - (60^\circ + 30^\circ) = 90^\circ$.
Because the ray strikes face AC perfectly at $90^\circ$ (perpendicular to the actual surface), its angle of incidence relative to the surface normal is exactly $0^\circ$.
Consequently, the ray completely emerges out of the prism from face AC perfectly undeviated, traveling in a straight line out into the air.
