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 ${35}^0$, suffers a deviation of ${40}^0$ and that it emerges at angle ${79}^0$. In which of the following is closest to the maximum possible value of the refractive index ?

• A. 1.8
• B. 1.5
• C. 1.6
• D. 1.7

Answer 1

The correct option is B 1.5

Given, $i={35}^{\circ }$, $\delta ={40}^{\circ }$, and $r={79}^{\circ }$.

We know, minimum deviation for a prism is,

$\delta =i+r-A=({35}^{\circ }+{79}^{\circ })-A$

$\therefore A={74}^{\circ }$

Now, maximum possible value of refractive index will be,

${\mu }_{max}=\frac{sin\left(\frac{A+\delta }{2}\right)}{sin\left(\frac{A}{2}\right)}=$

Therefore, ${\mu }_{max}=\frac{sin\left(\frac{{74}^{\circ }+{40}^{\circ }}{2}\right)}{sin\left(\frac{{74}^{\circ }}{2}\right)}$

Or, ${\mu }_{max}=\frac{sin{57}^{\circ }}{sin{37}^{\circ }}=1.44\approx 1.5$

So, the correct answer is - (B) $1.5$