Question

What is the minimum thickness of a thin film required for constructive interference in the reflected light from it given the refractive index of the film $ = 1.5\,nm$ , wavelength of the light incident on the film $ = 600\,nm$
A. $100\,nm$
B. $300\,nm$
C. $50\,nm$
D. $200\,nm$

Answer 1

Hint: we see thin film interference when light waves reflected off a film's top surface meet with waves reflected from the bottom surface. use the formula of constructive interference for a thin film. For minimum thickness $n$ will be zero. Put the values of refractive index and wavelength given and find the minimum thickness of the thin film

Formula Used:
condition for constructive interference $2\mu t = \left( {2n + 1} \right)\dfrac{\lambda }{2}$
Where $n = 0,1,2,3.....$, $\mu $ is the refractive index and $\lambda $ is the wavelength of the incident light.

Complete step by step answer:
When a lightwave is reflected off two surfaces at a distance equal to its wavelength, thin-film interference occurs. When light waves reflecting off the top and bottom surfaces collide, we witness a variety of coloured patterns. When light hits the boundary between two media, it is reflected in part and transmitted in part. The two reflections occur near each other when the second medium is a thin film.

As a result, two waves emerge from a thin film: one reflected off the film's top surface and the other reflected off the bottom surface. The phase difference between the two waves must be an even integral multiple of $\pi $ or ${180^0}$ for constructive interference to occur. The phase difference between the two waves in destructive interference is an odd integral multiple of $\pi $ or ${180^0}$

For constructive interference, we know
$2\mu t = \left( {2n + 1} \right)\dfrac{\lambda }{2}$
Where $n = 0,1,2,3.....$
$\mu $ is the refractive index
$\lambda $ is wavelength of the incident light
For minimum thickness we know , $n = 0$
$ \Rightarrow 2\mu t = \dfrac{\lambda }{2}$
$ \Rightarrow t = \dfrac{\lambda }{{4\mu }}$
$ \Rightarrow t = \dfrac{{600 \times {{10}^{^{ - 9}}}}}{{4 \times 1.5}}$
$ \therefore t = 100\,nm$

Hence option (A) $100\,nm$ is the right option.

Note: Constructive interference occurs when the maxima of the two waves, in the same phase, add up and the amplitude of the resultant wave is equal to the sum of the individual waves. On the other hand, destructive interference is when the minima of the two waves, in opposite phases, add up and the amplitude of the resultant wave is equal to the difference of the amplitudes of the two waves.