Question

A ray of light passes symmetrically through on equilateral prism of refractive index 1.5. determine

Answer 1

Answer:

$Angle \: at \: which \: minimum \: deviation \: occur = 37.18\degree$

Explanation:

Given:

Refractive index of prism μ = 1.5

Since the prism is equilateral,

Angle of prism A= 60°

To find:

Angel of incidence for which minimum deviation can occur(δm) ?

Solution:

We know that,

The refractive index, Angle of prism and angle of minimum deviation are correlated with formula,

$\mu \: = \frac{sin( \frac{A + \delta_m}{2} )}{sin \frac{A}{2} }$

Where symbols have their usual meaning.

Substituting the value from given data in formula,

$\mu \: = \frac{sin( \frac{A + \delta_m}{2} )}{sin \frac{A}{2} } \\1.5\: = \frac{sin( \frac{60 + \delta_m}{2} )}{sin \frac{60}{2} } \\ {sin( \frac{60 + \delta_m}{2} )} = 1.5 \times sin30 \\ {sin( \frac{60 + \delta_m}{2} )} = 1.5 \times 0.5 \\ {sin( \frac{60 + \delta_m}{2} )} = 0.75 \\ \frac{60 + \delta_m}{2} = {sin}^{ - 1}(0.75) \\ \frac{60 + \delta_m}{2} = 48.59 \: \degree \\ \delta_m = (48.59 \times 2) - 60\degree \\ \delta_m = (48.59 \times 2) - 60\degree \\ \delta_m = (97.18 - 60)\degree \\ \delta_m = 37.18\degree$

$Angle \: at \: which \: minimum \: deviation \: occur = 37.18\degree$

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