Question

At what angle should a ray of light be incident on the face of a prism of refracting angle $60^{\circ }$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$.

• A. $16^{\circ }$
• B. $29^{\circ }$
• C. $45^{\circ }$
• D. $58^{\circ }$

Answer 1

Answer: (B)

Angle of prism, $A=60^{\circ }$
Refractive index of prism $\mu =1.524$
Let $i$ be the angle of incidence. The critical angle is $i_C$ because it just suffers total internal refraction, so we use critical angle.
$\sin i_c=\frac{1}{\mu }=\frac{1}{1.524}=0.6561$
$i_c=41^{\circ }$
For a prism $r_1+r_2=A$ here $r_2=i_c$
$\therefore r_1+i_c=A$
$r_1+41^{\circ }=60^{\circ }$
$\Rightarrow r_1=19^{\circ }$
Using the formula, $\mu =\frac{\sin i_1}{\sin r_1}$
or $\sin i_1=1.524\sin 19^{\circ }=1.524\times 0.3256$
Or $i_1={\sin }^{-1}(0.4962)$
$i_1=29^{\circ }75^{\prime }$
Thus, the angle should be $29^{\circ }75^{\prime }$.