1. Use prism deviation relation

For a prism, $\delta = i + e - A$ where:

• $i$ = angle of incidence
• $e$ = angle of emergence
• $A$ = prism angle
• $\delta$ = deviation

Given: $i = 35^\circ, \quad \delta = 40^\circ, \quad e = 79^\circ$

So, $A = i + e - \delta = 35^\circ + 79^\circ - 40^\circ = 74^\circ$

2. Find the maximum possible refractive index

For a prism in air, the refractive index is $\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ when deviation is minimum.
But the question asks for the maximum possible refractive index for this prism angle.

The maximum value occurs when the ray just emerges grazing the second face, i.e. $e = 90^\circ$ Then inside the prism at second face, the internal angle of incidence equals the critical angle $C$: $r_2 = C$ Also, $r_1 + r_2 = A$ At the first face, $\sin i = \mu \sin r_1$ And at the limiting condition for emergence, $\sin C = \frac{1}{\mu}$

Thus, $r_2 = C = \sin^{-1}\left(\frac{1}{\mu}\right), \qquad r_1 = A - C$

Using Snell's law at first face, $\sin i = \mu \sin(A-C)$ with $i = 35^\circ$ and $A = 74^\circ$.

This is cumbersome directly, so use the equivalent limiting condition: At maximum possible refractive index for given $i$ and $A$, $r_1 = \sin^{-1}\left(\frac{\sin i}{\mu}\right), \qquad r_2 = \sin^{-1}\left(\frac{1}{\mu}\right)$ with $r_1 + r_2 = A = 74^\circ$

So solve: $\sin^{-1}\left(\frac{\sin 35^\circ}{\mu}\right) + \sin^{-1}\left(\frac{1}{\mu}\right) = 74^\circ$

3. Check options

Option C: $\mu = 1.5$

$\frac{\sin 35^\circ}{1.5} = \frac{0.574}{1.5} \approx 0.383$ $r_1 \approx \sin^{-1}(0.383) \approx 22.5^\circ$ $r_2 = \sin^{-1}\left(\frac{1}{1.5}\right) = \sin^{-1}(0.667) \approx 41.8^\circ$ $r_1 + r_2 \approx 22.5^\circ + 41.8^\circ = 64.3^\circ$ Not enough.

Option D: $\mu = 1.6$

$\frac{0.574}{1.6} \approx 0.359$ $r_1 \approx 21.0^\circ$ $r_2 = \sin^{-1}(0.625) \approx 38.7^\circ$ $r_1 + r_2 \approx 59.7^\circ$ Still not $74^\circ$.

As $\mu$ increases, both $r_1$ and $r_2$ decrease, so $r_1+r_2$ decreases. Hence these values are too large to satisfy the exact limiting equation? Let's instead derive properly from the given actual ray.

4. Use the actual ray data to compute refractive index

Since $A = 74^\circ$ Let internal refraction angles be $r_1, r_2$. Then $r_1 + r_2 = 74^\circ$

At first face: $\sin 35^\circ = \mu \sin r_1$

At second face: $\mu \sin r_2 = \sin 79^\circ$

So, $\sin r_1 = \frac{\sin 35^\circ}{\mu}, \qquad \sin r_2 = \frac{\sin 79^\circ}{\mu}$

Thus, $\sin^{-1}\left(\frac{\sin 35^\circ}{\mu}\right)+\sin^{-1}\left(\frac{\sin 79^\circ}{\mu}\right)=74^\circ$

Now test options.

Option C: $\mu=1.5$

$\frac{\sin35^\circ}{1.5} = \frac{0.574}{1.5}=0.383 \Rightarrow r_1\approx22.5^\circ$ $\frac{\sin79^\circ}{1.5} = \frac{0.982}{1.5}=0.655 \Rightarrow r_2\approx40.9^\circ$ $r_1+r_2\approx63.4^\circ$ Not enough.

Option D: $\mu=1.4$ would give bigger sum, so actual $\mu$ must be smaller than $1.5$.

Let us estimate:

For $\mu=1.33$, $\frac{0.574}{1.33}\approx0.432 \Rightarrow r_1\approx25.6^\circ$ $\frac{0.982}{1.33}\approx0.738 \Rightarrow r_2\approx47.5^\circ$ $r_1+r_2\approx73.1^\circ$ Very close.

So actual refractive index is about $\mu \approx 1.32\text{ to }1.33$

5. Interpret “maximum possible value”

Given the measured data, the refractive index comes out to be about $1.33$. Among the options $1.5,1.6,1.7,1.8$, the closest is $1.5$

So the nearest option is C.

6. Final answer

$\boxed{1.5}$ So, Option C is correct.