Question

In an experiment for determination of refractive index of glass of a prism by $i-\delta$ , plot, it was found that a ray incident at angle $35^{\circ }$, suffers a deviation of $40^{\circ }$ and that it emerges at angle $79^{\circ }$. Ιn that case which of the following is closest to the maximum possible value of the refractive index ?

• A. 1.5
• B. 1.6
• C. 1.7
• D. 1.8

Answer 1

Answer: (A)

$\delta =i+e-A$
$40=35+79-A$
$40=114-A$
$A=114-40=74=r_1+r_2$
From this we get,
$\mu =1.5$
$\therefore {\delta }_{min}<40^{\circ }$
$\mu <\frac{\sin \left(\frac{70+40}{2}\right)}{\sin 37}$
$\therefore {\mu }_{max}=1.44$