Question

When a thin transparent sheet of refractive index µ = 1.5 is placed near one of the slits in Young’s double-slit experiment; the intensity at the centre of the screen reduces to half of the maximum intensity. The minimum thickness of the sheet should be ( \[\lambda \]is the wavelength of light and both slits are identical)
A. \[\dfrac{\lambda }{4}\]
B. \[\dfrac{\lambda }{8}\]
C. \[\dfrac{\lambda }{2}\]
D. \[\dfrac{\lambda }{3}\]

Answer 1

Hint: In Young’s double-slit experiment interference of light takes place and due to constructive and destructive interference of light, we get the pattern on the screen.

Complete step by step answer:
 It is given that when a transparent sheet of refractive index µ = 1.5 is placed at one of the slits then intensity at the centre becomes one half.
We know that \[I=4{{I}_{{}^\circ }}{{\cos }^{2}}\dfrac{\phi }{2}\]
From given condition \[I=2{{I}_{{}^\circ }}\]
\[2{{I}_{{}^\circ }}=4{{I}_{{}^\circ }}{{\cos }^{2}}\dfrac{\phi }{2}\]
\[1=2{{\cos }^{2}}\dfrac{\phi }{2}\]
\[0.5={{\cos }^{2}}\dfrac{\phi }{2}\]
\[\phi =\dfrac{\pi }{2}\]
Now path difference \[\Delta x=\dfrac{\lambda }{2\pi }\times \phi \]
Substituting the values of \[\phi \]we get,
\[\begin{align}
  & =\dfrac{\lambda }{2\pi }\times \dfrac{\pi }{2} \\
 & =\dfrac{\lambda }{4} \\
\end{align}\]
\[\Delta x=\dfrac{\lambda }{4}\]
Now as per given conditions,
\[\begin{align}
  & (\mu -1)t=\dfrac{\lambda }{4} \\
 & (1.5-1)t=\dfrac{\lambda }{4} \\
 & 0.5t=\dfrac{\lambda }{4} \\
 & t=\dfrac{\lambda }{2} \\
 & \\
\end{align}\]

So, the correct answer is “Option C”.

Additional Information:
When an interference pattern is observed on the screen one distinct feature is all the bright bands are equally bright and all the dark bands are equally dark. While in case of diffraction when the pattern is observed on the screen the bright bands are of decreasing intensity as we move away from the central maxima. This is a very clear distinction between the two patterns.

Note:
Interference of light is a phenomenon in which two or more waves interfere with each other and there is redistribution of energy. We can say two waves superpose to form a resultant wave of greater, lower, or the same amplitude.