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

Concept: Since the ray is incident normally on the first face, \[ i_1=0^\circ. \] Therefore, \[ r_1=0^\circ. \] The ray travels along the normal to the first face. For a prism, \[ r_1+r_2=A. \] Hence, \[ r_2=A. \]

Step 1: Analyze the emergence from the second face. From the figure, the ray emerges grazing the second face. Therefore the angle of refraction is \[ e=90^\circ. \] Hence the internal angle of incidence at the second face equals the critical angle \(C\). \[ A=C. \]

Step 2: Find the critical angle. \[ \sin C = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3}. \] Thus, \[ C = \sin^{-1}\!\left(\frac{2}{3}\right). \] Since \[ A=C, \] \[ A = \sin^{-1}\!\left(\frac{2}{3}\right). \] \[\begin{aligned} \boxed{ A=\sin^{-1}\!\left(\frac{2}{3}\right) } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.