question_answer

In Young's double slit experiment, the introduction of a thin transparent film reduces the intensity at centre of screen by 75%. Then (\[\mu \]= refractive index of film, t = thickness of film and \[\lambda \] = wavelength of light used)

A) \[\mu =\frac{5}{2}\,\,if\,\,\lambda =2t\]

B) \[\mu =\frac{5}{3}\,\,if\,\,\lambda =t\]

C) \[\mu =\frac{5}{3}\,\,if\,\,\lambda =3t\]

D) \[\mu =\frac{4}{3}for\,\,any\,\,value\,\,of\,\,\lambda \]

Correct Answer: B

Solution :

[B]

As intensity decrease by 75% i.e. intensity is 25% of max intensity

i.e. I at center of screen \[=\frac{4{{I}_{0}}}{4}={{I}_{0}}\]

\[I={{I}_{0}}+{{I}_{0}}+2\sqrt{{{I}_{0}}}\sqrt{{{I}_{0}}}\cos \phi \]

\[\frac{4{{I}_{0}}}{4}=2{{I}_{0}}+2{{I}_{0}}\cos \phi \]

\[\therefore \cos \phi =-\frac{1}{2}\Rightarrow \phi =\frac{2\pi }{3},\frac{4\pi }{3}........\]

and \[\phi =\frac{2\pi }{\lambda }\Delta x\]

\[\Delta {{x}_{at}}o=(\mu -1)t\]

\[\phi =\frac{2\pi }{\lambda }(\mu -1)t\]

\[\phi =\frac{4\pi }{3}\]

\[\frac{4\pi }{3}=\frac{2\pi }{\lambda }(\mu -1)t\]

Put \[t=\lambda \]

\[\frac{2}{3}=\left( \frac{\mu -1}{\lambda } \right)\lambda \]

\[\mu =\frac{5}{3}\]