Question:
In the ideal double-slit experiment, when a glass-plate
(refractive index 1.5) of thickness t is introduced in the path
of one of the interfering beams (wavelength $\lambda$), the intensity
at the position where the central maximum occurred
previously remains unchanged. The minimum thickness of
the glass-plate is
In the ideal double-slit experiment, when a glass-plate
(refractive index 1.5) of thickness t is introduced in the path
of one of the interfering beams (wavelength $\lambda$), the intensity
at the position where the central maximum occurred
previously remains unchanged. The minimum thickness of
the glass-plate is
Updated On: May 19, 2022
- 2$\lambda$
- $\frac{2 \lambda}{3}$
- $\frac{\lambda}{3}$
- $\lambda$
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The Correct Option is A
Solution and Explanation
Path difference due to slab should be integral multiple of $\lambda$ or
$\hspace10mm ? x=n \lambda \, or \, (\mu-1)t=n \lambda$ n = 1,2,3,..
or $\hspace20mm t=\frac{n \lambda}{\mu-1}$
For minimum value of t, n = 1
$\therefore \hspace20mm t=\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}=2 \lambda$
$\hspace10mm ? x=n \lambda \, or \, (\mu-1)t=n \lambda$ n = 1,2,3,..
or $\hspace20mm t=\frac{n \lambda}{\mu-1}$
For minimum value of t, n = 1
$\therefore \hspace20mm t=\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}=2 \lambda$
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