In an experiment for determination of refractive index of glass of a prism by i v/s $$\delta$$ 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?

First, we recall the two standard relations that hold for a prism of refracting angle $$A$$ and refractive index $$\mu$$:

1. The deviation produced for a given pair of angles of incidence $$i$$ and emergence $$e$$ is

$$\delta=i+e-A.$$

2. When the deviation becomes minimum (that special value is denoted by $$\delta_m$$) the path of the ray inside the prism is symmetric, so $$i=e=i_m$$ and

$$\mu=\dfrac{\sin\!\left(\dfrac{A+\delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}.$$

In the experiment one particular ray is reported to have

$$i=35^\circ,\qquad \delta=40^\circ,\qquad e=79^\circ.$$

Using the first formula we find the prism angle $$A$$:

$$\begin{aligned} \delta &= i+e-A\\[4pt] 40^\circ &= 35^\circ+79^\circ-A\\[4pt] A &= 35^\circ+79^\circ-40^\circ\\[4pt] A &= 74^\circ. \end{aligned}$$

The experiment has provided only one point on the $$i$$ versus $$\delta$$ graph, so the actual minimum deviation $$\delta_m$$ is not known exactly. However, the deviation curve is always U-shaped; hence the measured deviation of $$40^\circ$$ can only be greater than or equal to the minimum deviation. Symbolically,

$$\delta_m\le 40^\circ.$$

The refractive index obtained from the second formula grows monotonically with $$\delta_m$$ because the numerator $$\sin\!\bigl(\tfrac{A+\delta_m}{2}\bigr)$$ increases as $$\delta_m$$ increases while the denominator is fixed. Therefore, the largest possible value of $$\mu$$ will correspond to the largest admissible value of $$\delta_m$$, namely

$$\delta_m^{\text{(max)}}=40^\circ.$$

Substituting $$A=74^\circ$$ and $$\delta_m=40^\circ$$ in the second relation gives

$$\begin{aligned} \mu_{\text{max}}&=\dfrac{\sin\!\left(\dfrac{A+\delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}\\[6pt] &=\dfrac{\sin\!\left(\dfrac{74^\circ+40^\circ}{2}\right)}{\sin\!\left(\dfrac{74^\circ}{2}\right)}\\[6pt] &=\dfrac{\sin(57^\circ)}{\sin(37^\circ)}.\\[6pt] \end{aligned}$$

Now we evaluate the sines step by step:

$$\sin(57^\circ)\approx0.8387,\qquad\sin(37^\circ)\approx0.6018,$$

so

$$\mu_{\text{max}}\approx\dfrac{0.8387}{0.6018}\approx1.39.$$

The numerical value $$1.39$$ is not one of the quoted options. We must therefore look for the option that is closest to it. Comparing with the given choices

$$1.5,\;1.6,\;1.7,\;1.8,$$

we see that $$1.5$$ is the nearest to $$1.39$$.

Hence, the correct answer is Option C.