Animated Solution for Physics - Optics: A ray of light $AO$ in vacuum is incident on a glass slab at angle $60^{\circ }$ and refracted at angle $30^{\circ }$ along $OB$ as shown in the figure. The optical path length of light ray from $A$ to $B$ is

The journey of light through different media is one of the most fascinating phenomena in physics. When light travels from a vacuum into a denser medium like glass, it doesn't just change direction; its effective "path length" changes as well. This problem is a beautiful amalgamation of basic geometry, Snell's Law, and the profound concept of optical path length. Let's dive deep into the mechanics of this problem and unravel the solution step by step.

A Tale of Two Media

Imagine a ray of light, which we will call $AO$, traveling through the emptiness of a vacuum. It strikes the smooth surface of a glass slab at an angle of incidence of $60^{\circ }$. The moment it hits the glass at point $O$, it slows down and bends towards the normal, refracting at an angle of $30^{\circ }$ along the path $OB$.

We are given the vertical distances for both segments of the journey: the vertical drop from point $A$ to the interface is $a$, and the vertical drop from the interface to point $B$ is $b$. Our ultimate goal is to find the optical path length from $A$ to $B$.

But what exactly is optical path length? It is not just the physical distance you would measure with a ruler. The optical path length represents the distance light would have traveled in a vacuum during the same time it took to travel through the given medium. Mathematically, it is defined as the product of the geometric path length and the refractive index of the medium.

For our specific problem, the total optical path length is the sum of the optical path in the vacuum and the optical path in the glass. Since the refractive index of a vacuum is exactly $1$, the optical path for the first segment is simply the geometric length $AO$. For the second segment, it is the geometric length $OB$ multiplied by the refractive index of the glass, which we will call $\mu$.

Therefore, our master equation becomes:

To use our master equation, we first need to find the actual physical lengths of the paths $AO$ and $OB$. This is where our given vertical distances $a$ and $b$ come into play.

Let's focus on the first segment, the ray $AO$ in the vacuum. If we draw a vertical line down from $A$ to the horizontal interface, we form a right-angled triangle. The angle between the ray $AO$ and the vertical normal is $60^{\circ }$. In this triangle, the side adjacent to the $60^{\circ }$ angle is our vertical distance $a$, and the hypotenuse is the path length $AO$.

Using basic trigonometry, we can relate these quantities using the cosine function:

We know that $\cos 60^{\circ }$ is equal to $\frac{1}{2}$. Substituting this value, we can solve for $AO$:

Now, let's apply the exact same logic to the refracted ray $OB$ inside the glass. We have another right-angled triangle formed by the ray $OB$, the normal, and the vertical distance $b$. The angle of refraction is $30^{\circ }$.

Again, using the cosine function:

We know that $\cos 30^{\circ }$ is equal to $\frac{\sqrt{3}}{2}$. Substituting this value, we can solve for $BO$:

We have successfully found the geometric lengths $AO$ and $OB$. However, to calculate the optical path length, we still need one crucial piece of the puzzle: the refractive index of the glass, $\mu$.

This is where Snell's Law comes to our rescue. Snell's Law elegantly relates the angles of incidence and refraction to the refractive indices of the two media. For light traveling from a vacuum (where the refractive index is $1$) into a medium with refractive index $\mu$, Snell's Law is stated as:

We are given the angle of incidence $i=60^{\circ }$ and the angle of refraction $r=30^{\circ }$. Let's substitute these values into the equation:

Now, we plug in the standard trigonometric values. We know that $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ and $\sin 30^{\circ }=\frac{1}{2}$:

Notice how the denominators perfectly cancel each other out. This leaves us with a beautifully simple result for the refractive index:

We now have all the ingredients required to calculate the final optical path length. Let's recall our master equation:

We have determined that: 1. $AO=2a$ 2. $BO=\frac{2b}{\sqrt{3}}$ 3. $\mu =\sqrt{3}$

Let's carefully substitute these values back into the master equation:

Take a close look at the second term. The $\sqrt{3}$ in the numerator (from the refractive index) and the $\sqrt{3}$ in the denominator (from the geometric length $BO$) perfectly cancel each other out. This is a hallmark of a well-designed physics problem!

After the cancellation, we are left with a remarkably clean and elegant final expression:

This result tells us that the effective optical distance the light travels is simply twice the sum of the vertical drops. By systematically breaking down the problem into its geometric and physical components, we have arrived at the correct answer, which corresponds to option (c).