Discussion Forum

In an experiment for determination of the refractive index of a glass of a prism by i-δ plot, ut was found that a ray incident at an angle 35° suffers a deviation of 40° and that it emerges at an angle 79⁰, In that case, which of the following is closest to the maximum possible value of the refractive index?

Answer:

Explanation:

$\delta = (i_{1}+i_{2})-A$

   $\Rightarrow 40^{0}=(35^{0}+79^{0})-A$

 $\Rightarrow  A=74^{0}$

Now, we know that  $\mu = \frac{\sin(\frac{A+\delta_{m}}{2})}{\sin(\frac{A}{2})}$

It  we take the given deviation as the minimum  deviation then,

$\mu = \frac{\sin(\frac{74^{0}+40^{0}}{2})}{\sin(\frac{74^{0}}{2})}$ = 1.51

The given deviation may or may not be the minimum deviation. Rather it will be less than this value. Therefore  μ  will be less than 1.51

  Hence , the maximum possible value of refractive index is 1.51