Animated Solution for Physics - Optics: A prism of angle $A=1^{\circ }$ has a refractive index $\mu =1.5$. A good estimate for the minimum angle of deviation (in degree) is close to $N/10$. Value of $N$ is ……… .

Light is a fascinating entity. It travels through the vacuum of space at an unimaginable speed, completely undisturbed. But when it encounters a dense medium—like a piece of glass—its journey changes. It slows down, and if it enters at an angle, it bends. This bending of light is the fundamental principle behind lenses, microscopes, and the classic optical element: the prism.

When a beam of light passes through a prism, it undergoes refraction twice—once upon entering and once upon exiting. The total change in the direction of the light ray is known as the angle of deviation, denoted by the Greek letter $\delta$.

For a standard prism, calculating this deviation requires navigating the complex trigonometric waters of Snell's Law at both interfaces. The deviation depends heavily on the angle at which the light strikes the first surface (the angle of incidence).

If you were to slowly rotate a prism while shining a laser through it, you would notice the deviated spot on the wall moving. At a very specific angle, the spot reaches a turning point—it stops moving in one direction and starts moving back. This unique point is called the angle of minimum deviation (${\delta }_m$). At this precise moment, the light ray travels symmetrically through the prism, parallel to its base.

The general formula connecting the refractive index $\mu$, the prism angle $A$, and the minimum deviation ${\delta }_m$ is:

While the general formula is universally correct, nature offers us a beautiful mathematical shortcut when the prism is extremely thin.

In our specific problem, we are given a prism with an angle $A=1^{\circ }$. This is definitely a thin prism. When angles are very small (typically less than $10^{\circ }$), the sine of the angle is approximately equal to the angle itself when measured in radians. This is known as the small-angle approximation: $\sin (\theta )\approx \theta$.

By applying this approximation to the standard prism formula, the complex sines melt away:

This leaves us with an incredibly elegant and linear relationship. Notice how the angle of incidence doesn't even appear in this final formula! For a thin prism, the deviation is practically constant regardless of how the light enters it, meaning the deviation is always essentially the minimum deviation.

Let's bring our simplified formula to the stage and substitute the known values. We are given the refractive index of the glass, $\mu =1.5$, and the prism angle $A=1^{\circ }$.

The math here is beautifully simple. Subtracting one from one point five leaves us with zero point five.

So, the light beam is deviated by exactly half a degree from its original path.

Solving for N

The problem presents us with a slight twist at the end. It states that the deviation is close to the expression $\frac{N}{10}$, and asks us to find the integer $N$.

We simply equate our calculated deviation to this expression:

To isolate $N$, we multiply both sides by ten.

And there we have it! The value of $N$ is exactly 5. What started as a potentially complex optics problem was elegantly unraveled using the power of mathematical approximations. Physics is not just about memorizing formulas; it's about knowing when and how to simplify reality to reveal the beautiful truths hidden beneath.