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Ydse With Thin Slab MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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YDSE with thin slab is considered one of the most asked concept.
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On introducing a thin mica sheet of thickness $2 \times 10^{-6} \mathrm{~m}$ and refractive index of 1.5 in the path of one of the waves, central bright maxima shifts by $\mathrm{n}$ fringes. The wavelength of the wave used is $5000 \AA$ , then $\mathrm{n}$ is

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In Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness $1.964 \mathrm{~mm}$ is introduced in the path of one of the two waves. If now the mica sheet is removed and the distance between the slit and screen is doubled, the distance between successive max. or min. remains unchanged. The wavelength of the monochromatic light used in the experiment is

$4000 \AA$

$5500 \AA$

$5892 \AA$

$6071 \AA$

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Concepts Covered - 1

YDSE with thin slab

YDSE with thin slab:

Consider the arrangement of Young's double slit experiment as shown in fig. In which a thin transparent film of refractive index $\mu$ and thickness 't' is introduced in front of the lower slit 'S'. Our aim is to obtain the new position of the neth maxima and minima . Let us assume a point P on screen at a distance Y from the origin O. It is important to note that in this particular situation, we cannot calculate the phase difference between the two waves arriving at $P$ directly by calculating the path difference $\left(S_{2} P-S_{1} P\right)$ because the two waves are not traveling in the same medium. The lower wave travels some distance in a medium $( \mu)$ and the remaining distance in air, while the upper wave travels all the distance in air andce traveled in the effective path difference we need to convert the distance traveled in medium $( \mu)$ into its equivalent distance air, which is equal to $( \mu)$ tand it is called the optical path. Hence optical path is the equivalent distance to be traveled in air to produce the same phase change as that produced in actual in traveling the actual distance. Thus, the optical path difference between the two waves is

$\mathrm{\Delta x}=\left[\left(\mathrm{S}_{2} \mathrm{P}-\mathrm{t}\right)+\mu \mathrm{t}\right]-\mathrm{S}_{1} \mathrm{P}$ or

$\mathrm{\Delta x}=\left(\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}\right)+(\mu-1) \mathrm{t}$

$\begin{array}{ll}{\text { since }} & {\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=\mathrm{d} \sin \theta=\mathrm{d}\left(\mathrm{y}^{\prime} / \mathrm{D}\right) \quad \text { (from the fig.) }} \\ {\therefore \quad} & {\mathrm{\Delta x}=\mathrm{d} \mathrm{y}_{\mathrm{n}}^{\prime} / \mathrm{D}+(\mu-1) \mathrm{t}}\end{array}$

From the nth maxima,

$\mathrm{\Delta x}=\mathrm{n} \lambda, \therefore \mathrm{n} \lambda=\mathrm{dy}_{\mathrm{n}} / \mathrm{D}+(\mu-1) \mathrm{t} \ \text{or}$

$y_n=\frac{n\lambda D}{d} - \frac{(\mu -1)tD}{d}$

The position of nth maxima and minima has shifted downward by the same
distance which is called $\mathrm{S}=\mathrm{y}_{\mathrm{n}}-\mathrm{y}_{\mathrm{n}}^{\prime}=(\mu-1) \frac{\mathrm{tD}}{\mathrm{d}}$

The distance between two successive maxima or minima remains unchanged. That is, the fringe width remains unchanged by introducing a transparent film.
The distance of shift is in the direction where the film is introduced. That is, if a film is placed in front of the upper slit the $S_{1},$ fringe pattern shifts upwards, if a film is placed in front of the lower slit $S_{2},$ the fringe pattern
shifts downward.