A ray of light passes through an equilateral prism such that an angle of incidence is equal to the angle of emergence and the latter is equal to (3/4) th the angle of prism. The angle of deviation is:

Question

A ray of light passes through an equilateral prism such that an angle of incidence is equal to the angle of emergence and the latter is equal to (3/4) th the angle of prism. The angle of deviation is: (A)  45°  (B)  39°  (C)  20°  (D)  30°

Tardigrade Admin · Accepted Answer

Angle of incidence = angle of emergence, i.e., i = i' Also, i' = (3/4)× angle of equilateral prism =(3/4)× 60° =45° Thus, angle of deviation = i + i' - A =(45° +45° -60° )=30°

Q. A ray of light passes through an equilateral prism such that an angle of incidence is equal to the angle of emergence and the latter is equal to $\frac{3}{4}$ th the angle of prism. The angle of deviation is:

$45^{\circ}$

$39^{\circ}$

$20^{\circ}$

$30^{\circ}$

$90^{\circ}$

Angle of incidence = angle of emergence, i.e., $i = i^{'}$
Also, $i^{'} = \frac{3}{4} \times$ angle of equilateral prism $= \frac{3}{4} \times 60^{\circ} = 45^{\circ}$
Thus, angle of deviation $= i + i^{'} - A$
$= (45^{\circ} + 45^{\circ} - 60^{\circ}) = 30^{\circ}$

Q. A ray of light passes through an equilateral prism such that an angle of incidence is equal to the angle of emergence and the latter is equal to 43​ th the angle of prism. The angle of deviation is:

45∘

39∘

20∘

30∘

90∘

Angle of incidence = angle of emergence, i.e., i=i′ Also, i′=43​× angle of equilateral prism =43​×60∘=45∘ Thus, angle of deviation =i+i′−A =(45∘+45∘−60∘)=30∘