Find the maximum angle of deviation for a prism with angle A = 60° and $μ$ = 1.5.
Answer & explanation
Correct answer: option 2
Maximum Deviation
The deviation is maximum when i = 90° or e = 90° that is at grazing incidence or grazing emergence.
Let i = 90°
$⇒ r_1 = C = \sin^{−1}(1/μ)$
$⇒ r_1 = \sin^{−1}(2/3) = 42°$
$⇒ r_2 = A −r_1 = 60° − 42° = 18°$
$μ \sin g\frac{\sin r_2}{\sin e}=\frac{1}{μ}$
$\sin e = u\, \sin r_2 = 1.5\, \sin 18°$
$⇒ \sin e= 0.463$
$⇒ r = 28°$
Deviation = $D_{max}$
$= i + e – A = 90° + 28° – 60° = 58°$