Wave Optics

Classical Mechanics Level 3

A glass plate of refractive index $1.5$ is coated with a thin layer of thickness $t$ and refractive index $1.8$ . A beam of light of wavelength $648 nm$ travelling in air is incident normally on the layer. It is partially reflected off the upper and the lower surface of the layer, and the two reflected rays interfere.

Find the minimum value of $t$ in $nm$ for which the two reflected rays interfere constructively.

You can try more of my Questions here .

The answer is 90.

1 solution

Tanishq Varshney
Feb 24, 2015

Let $\mu_{f}$ be the refractive index of Film $=1.8$

The path difference between partly reflected at the upper and the lower surface of the layer is $\triangle x$

=> $\triangle x=2\mu_{f}tcosr+\frac{\lambda }{2}$ [ $t$ is thickness of film]

As the rays are incident normally $r=0$ or $cosr=1$ [where $r$ is angle of refraction ]

For constructive interference, $\triangle x=n\lambda$

=> $2\mu_{f}t+\frac{\lambda}{2}=n\lambda$

=> $2\mu_{f}t_{minimum}+\frac{\lambda}{2}=\lambda$ [ $t$ is minimum when $n=1$ ]

=> $t_{min}=\frac{\lambda}{4\mu_{f}}$

=> $t_{min}=\boxed{90}$

Nice solution $\ddot\smile$

A Former Brilliant Member - 6 years, 3 months ago

sir,in del(X), Y is there a lambda/2...Is it because the wave changes phase by 180 in refraction...can u please explain this...tnks

incredible mind - 6 years, 3 months ago

No that is due to reflection there is a change corresponding to half the wavelength and hence the justification. During reflection at an interface if we consider wave model then this condition is bound to occur

Sarthak Shiv - 3 years, 7 months ago

Wave Optics

The answer is 90.

1 solution

0 pending reports