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

The ray's path is analyzed step-by-step using geometrical optics: - Incident ray \( \text{MN} \) strikes surface AB perpendicularly. No refraction occurs, and the ray enters the prism without deviation.

- Prism \( \triangle ABC \) is a right-angled isosceles prism with angles \( \angle A = 90^\circ \), \( \angle B = 45^\circ \), and \( \angle C = 45^\circ \). The ray propagates towards surface BC. - At surface BC, the angle of incidence is \( i = 45^\circ \) (measured from the normal, due to the isosceles triangle). The critical angle for the glass-air interface is calculated: \[ \sin C = \frac{1}{\mu} = \frac{1}{1.6} \approx 0.625 \Rightarrow C \approx 38.68^\circ \] Since \( i = 45^\circ>38.68^\circ \), total internal reflection (TIR) occurs at face BC. - The ray reflects from BC towards face AC, which is adjacent to the second prism \( \triangle DBC \). As both prisms are of the same glass and this interface is not with air, no refraction occurs at the BC interface between the two prisms. - Prism \( \triangle DBC \) is an equilateral triangle. At face CD, the ray strikes at an interior angle of \( 60^\circ \). TIR is checked again: \[ i = 60^\circ>C = 38.68^\circ \Rightarrow \text{TIR occurs at CD as well} \] - Finally, the ray emerges normally from face AD, striking it perpendicularly after the internal reflections. Path of the ray: MN → undeviated into the prism → reflects from BC → reflects from CD → exits normally through AD.