Question

A right-angled isosceles glass prism ABC is kept in contact with an equilateral triangular prism DBC as shown in the figure. Both prisms are made of the same glass of refractive index 1.6. Trace the path of the ray MN incident normally on face AB as it passes through the combination.

Hint:

Check for total internal reflection using critical angle: \( \sin C = \frac{1}{\mu} \). If incidence angle exceeds critical angle at a glass-air interface, TIR will occur.

Answer 1

Let us analyze the passage of the ray step-by-step using geometrical optics: - The ray \( \text{MN} \) is incident normally on the surface AB, which means there is no refraction at that surface. The ray enters the prism undeviated.

- The prism \( \triangle ABC \) is a right-angled isosceles prism with \( \angle A = 90^\circ \), \( \angle B = 45^\circ \), and \( \angle C = 45^\circ \). The ray inside travels toward surface BC. - At surface BC, the ray strikes at an angle of incidence \( i = 45^\circ \) (measured from the normal, because triangle is isosceles). Let’s calculate the critical angle for glass-air interface: \[ \sin C = \frac{1}{\mu} = \frac{1}{1.6} \approx 0.625 \Rightarrow C \approx 38.68^\circ \] Since \( i = 45^\circ>38.68^\circ \), total internal reflection (TIR) occurs at face BC. - The ray reflects from BC toward face AC, which is in contact with the second prism \( \triangle DBC \). Since both prisms are made of the same glass, and this surface is not in contact with air, no refraction occurs at BC interface between the two prisms. - Now, consider triangle \( \triangle DBC \) (an equilateral triangle). At face CD, the ray strikes at \( 60^\circ \) (interior angle). Again, let’s check for TIR: \[ i = 60^\circ>C = 38.68^\circ \Rightarrow \text{TIR occurs at CD also} \] - Finally, the ray emerges normally out of face AD because it strikes it perpendicularly after internal reflections. Path of the ray: MN → undeviated into the prism → reflects from BC → reflects from CD → exits normally through AD.