Question

A ray of light is incident normally on a refracting face of a prism. The subsequent journey of the ray through the prism is shown in the figure. The refractive index of the prism material is \(1.5\). Find the angle of the prism \((\angle A)\).

• A. \[ \sin^{-1}\!\left(\frac{3}{2}\right) \]
• B. \[ \sin^{-1}\!\left(\frac{2}{3}\right) \]
• C. \[ \cos^{-1}\!\left(\frac{2}{3}\right) \]
• D. \[ \sin^{-1}\!\left(\frac{4}{3}\right) \]

Hint:

If a ray enters a prism normally, \[ r_1=0. \] Therefore, \[ A=r_2. \] When the emergent ray grazes the surface, \[ r_2=C, \] where \[ \sin C=\frac{1}{\mu}. \]

Answer 1

The Correct Option is B

Step 1: First face, normal incidence.
The ray strikes the first face straight on, so the angle of incidence there is $0^\circ$ and it passes in without bending. Inside, it travels along the normal to that first face.

Step 2: Prism angle relation.
For a prism the two refraction angles add to the prism angle: \[ r_1+r_2=A. \] Since $r_1=0$, we get $r_2=A$. So the angle at which the ray meets the second face is exactly the prism angle $A$.

Step 3: Read the second-face behaviour.
The figure shows the ray just grazing along the second face as it leaves, which means it hits that face at the critical angle for total internal reflection onset. So $r_2=\theta_c=A$.

Step 4: Critical-angle formula.
At the critical angle, \[ \sin\theta_c=\frac{1}{\mu}. \]

Step 5: Put in the refractive index.
With $\mu=1.5$, \[ \sin A=\frac{1}{1.5}=\frac{2}{3}. \]

Step 6: Solve for the prism angle.
\[ A=\sin^{-1}\!\left(\frac{2}{3}\right). \]
\[ \boxed{A=\sin^{-1}\!\left(\dfrac{2}{3}\right)} \]