Question

A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index of 1.8. Light of wavelength $\lambda $ travelling in air is incident normally on the layer. It is partly reflected at the upper and lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If $\lambda = 648nm$, obtain the least value of t for which the rays interfere constructively:
A) 90nm.
B) 40nm.
C) 30nm.
D) 45nm.

Answer 1

Hint:The constructive interference occurs when the two waves superimpose and the maximum of the two add together and so the amplitude of the two add together and then give the amplitude of the resulting wave.

Complete step by step answer:
It is given that a glass plate of refractive index of 1.5 is coated with a thin layer of thickness t and the refractive index of 1.8. Light of wavelength $\lambda $ travelling in air is incident normally on the layer. It is partly reflected at the upper and lower surfaces of the layer and the two reflected rays interfere.

We need to find the condition for constructive interference for least value of thickness t if the wavelength of the ray of light is $\lambda = 648nm$. When reflection happens then the change of phase takes place and the value of the change of phase is $\pi $. For the constructive interference the relation for the distance travelled by the ray will be equal to,
$\Delta x = \left( {2n - 1} \right) \cdot \dfrac{\lambda }{2}$………eq. (1)
Also the distance travelled by the rays can be expressed as,
$\Delta x = 2\mu t$
As the material has refractive index ${\mu _2}$ and the thickness is minimum therefore,
$ \Rightarrow \Delta x = 2{\mu _2}{t_{\min .}}$………eq. (2)
Equating equation (1) and equation (2) we get,
$ \Rightarrow 2{\mu _2}{t_{\min .}} = \left( {2n - 1} \right) \cdot \dfrac{\lambda }{2}$
Now for minimum thickness the value of n becomes$n = 1$.
$ \Rightarrow 2{\mu _2}{t_{\min .}} = \left( {2n - 1} \right) \cdot \dfrac{\lambda }{2}$
Replace the value of $n = 1$ in the relation.
$ \Rightarrow 2{\mu _2}{t_{\min .}} = \left( {2 - 1} \right) \cdot \dfrac{\lambda }{2}$
$ \Rightarrow 2{\mu _2}{t_{\min .}} = \dfrac{\lambda }{2}$
As it is given that the wavelength of the ray is $\lambda = 648nm$ and the refractive index of the thin layer is equal to${\mu _2} = 1 \cdot 8$. Therefore replace the value of wavelength and refractive index in the above relation.
$ \Rightarrow 2{\mu _2}{t_{\min .}} = \dfrac{\lambda }{2}$
Put $\lambda = 648nm$ and${\mu _2} = 1 \cdot 8$.
$ \Rightarrow 2\left( {1 \cdot 8} \right){t_{\min .}} = \dfrac{{648}}{2}$
$ \Rightarrow {t_{\min .}} = \dfrac{{648}}{{4 \times 1 \cdot 8}}$
After solving the value of ${t_{\min .}}$ we get,
$ \Rightarrow {t_{\min .}} = 90nm$
So the value of the minimum thickness is ${t_{\min .}} = 90nm$.

The correct answer for this problem is option A.

Note: When a wave is changing medium then some part of the wave passes through and some part get reflected the part of wave which gets reflected experiences change in the phase of the wave and the change of the phase is always $\pi $.