In an experiment for determination of refractive index of glass of a prism by i – δ, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index ?

Engineering

Physics

Prism

Question

1.7

1.6

1.8

1.5

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δ = 1 + e – A ⇒ A = 74°

$μ = \frac{\sin (\frac{a + δ_{\min}}{2})}{\sin (\frac{A}{2})} = \frac{5}{3} \sin (37 ° + \frac{δ_{\min}}{2})$

$μ_{\max} can be \frac{5}{3}, so μ will be less than \frac{5}{3}$

Since δ_min will be less then 40°. So

$μ < \frac{5}{3} \sin 57 ° < \frac{5}{3} \sin 60 ° \Rightarrow μ < 1 .446$

So the nearest possible value of μ should be 1.5