Answer:
[tex] Angle \: at \: which \: minimum \: deviation \: occur = 37.18\degree[/tex]
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,
[tex]\mu \: = \frac{sin( \frac{A + \delta_m}{2} )}{sin \frac{A}{2} } [/tex]
Where symbols have their usual meaning.
Substituting the value from given data in formula,
[tex] \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[/tex]
[tex] Angle \: at \: which \: minimum \: deviation \: occur = 37.18\degree[/tex]
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