Animated Solution for Physics - Optics: A glass plate of refractive index 1.5 is coated with a thin layer of thickness $t$ and refractive index 1.8. Light of wavelength $\lambda$ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If $\lambda =648\text{ nm}$, obtain the least value of $t$ for which the rays interfere constructively.

Imagine you are looking at a beautifully coated piece of glass. We have a glass plate with a refractive index of $1.5$, and on top of it sits a thin layer of a different material with a refractive index of $1.8$. Light from the air (where the refractive index is $1$) is shining directly down onto this setup.

When the light hits the first boundary (between the air and the thin layer), part of it reflects back up. We'll call this Ray 1. The rest of the light travels through the thin layer, hits the second boundary (between the thin layer and the glass), and reflects back up again. We'll call this Ray 2. These two reflected rays will eventually meet and interfere with each other. Our goal is to find the condition for them to interfere constructively, meaning they combine to form a bright spot.

To understand how these rays interfere, we first need to look at the optical path difference between them. Ray 2 travels an extra distance by going down through the thin layer of thickness $t$ and coming back up. Since it travels this distance inside a medium with a refractive index of ${\mu }_1=1.8$, the optical path difference is:

But wait, there is a catch here! We must also consider the phase changes that occur upon reflection. When light reflects off a medium that is optically denser (has a higher refractive index), it undergoes a phase shift of $\pi$. Ray 1 reflects off the thin layer (${\mu }_1=1.8$) while traveling in air ($\mu =1$). Since $1.8>1$, Ray 1 experiences a phase shift of $\pi$.

On the other hand, Ray 2 reflects off the glass (${\mu }_2=1.5$) while traveling in the thin layer (${\mu }_1=1.8$). Since $1.5<1.8$, it is reflecting off a rarer medium, so it experiences no phase shift.

Because only one of the rays experiences a phase shift of $\pi$, the two rays have an inherent phase difference of $\pi$. For them to interfere constructively, the optical path difference must compensate for this by being an odd multiple of half the wavelength. Therefore, the condition for constructive interference is:

Now, let's substitute our optical path difference into the condition:

We are asked to find the least value of the thickness $t$. To make $t$ as small as possible, we should choose the smallest valid integer for $n$, which is $n=1$. Substituting $n=1$ gives:

We are given that the wavelength of the light is $\lambda =648\text{ nm}$. Let's plug that in and solve for $t$:

And there we have it! The minimum thickness of the layer required for constructive interference is $90\text{ nm}$.