A prism of angle $$A = 1^\circ$$, $$\mu = 1.5$$. A good estimate for the minimum angle of deviation (in degrees) is close to $$\frac{N}{10}$$. Value of N is.........

We have a prism whose refracting angle is $$A = 1^\circ$$ and whose refractive index is $$\mu = 1.5$$. For a prism kept in the position of minimum deviation, the angle of incidence $$i$$ equals the angle of emergence $$e$$, and the ray inside the prism makes equal angles with both faces. Hence the refraction angle inside the prism is

$$r = \frac{A}{2}.$$

Substituting the given angle of the prism,

$$r = \frac{1^\circ}{2} = 0.5^\circ.$$

Now we invoke Snell’s law at the first refracting surface. The law states

$$\mu = \frac{\sin i}{\sin r}.$$

Rearranging it to obtain the incidence angle, we get

$$\sin i = \mu \sin r.$$

Substituting $$\mu = 1.5$$ and $$r = 0.5^\circ$$,

$$\sin i = 1.5 \,\sin 0.5^\circ.$$

Because the angle $$0.5^\circ$$ is very small, we may convert it first into radians to evaluate its sine accurately. We remember the relation

$$1^\circ = \frac{\pi}{180}\ \text{radian},$$

so

$$0.5^\circ = 0.5 \times \frac{\pi}{180} \text{ rad} = \frac{\pi}{360} \text{ rad} \approx 0.00872665 \text{ rad}.$$

Using the small-angle approximation $$\sin \theta \approx \theta$$ (valid when $$\theta$$ is expressed in radians),

$$\sin 0.5^\circ \approx 0.00872665.$$

Multiplying by $$1.5$$ gives

$$\sin i \approx 1.5 \times 0.00872665 = 0.0130899.$$

Again, since this value is small, $$i$$ itself is small, and we can use $$\sin^{-1} x \approx x$$ in radians. Thus

$$i \approx 0.0130899 \text{ rad}.$$

To express this in degrees we multiply by $$\frac{180}{\pi}$$:

$$i \approx 0.0130899 \times \frac{180}{\pi} = 0.0130899 \times 57.2958 \approx 0.75^\circ.$$

The minimum angle of deviation $$\delta_{\min}$$ for a prism is given by the formula

$$\delta_{\min} = 2i - A.$$

Substituting $$i = 0.75^\circ$$ and $$A = 1^\circ$$, we obtain

$$\delta_{\min} = 2 \times 0.75^\circ - 1^\circ = 1.5^\circ - 1^\circ = 0.5^\circ.$$

The problem statement says that $$\delta_{\min}$$ is “close to $$\dfrac{N}{10}^\circ$$”. Since $$0.5^\circ = \dfrac{5}{10}^\circ$$, we read off

$$N = 5.$$

So, the answer is $$5$$.