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Ydse With Thin Slab - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • YDSE with thin slab is considered one of the most asked concept.

  • 10 Questions around this concept.

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A thin mica sheet of thickness 4 \times 10^{-6} \mathrm{~m} and refractive index (\mu=1.5) is introduced in the path of the light from the upper slit. The wavelength of the wave used in \mathrm{ 5000\, \AA}. The central bright maximum will shift:

A thin mica sheet of thickness 2 \times 10^{-6} \mathrm{~m} and refractive index \mathrm{(\mu=1.5)} is introduced in the path of the light from upper slit. The wavelength of the wave used is \mathrm{5000\, \AA}. The central bright maximum will shift:

In Young's double slit experiment, the y-coordinates of central maxima and 10^{\text {th }} maxima are 2 \mathrm{~cm} and 5 \mathrm{~cm}, respectively. When the YDSE apparatus is immersed in a liquid with a refractive index of 1.5. The corresponding y coordinates will be:

A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected in any reflection by upper surface and 50 % of light is reflected by any reflection from lower surface. Rest is refracted. The ratio of maximum to minimum intensity in interference region of reflected rays is:

 

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.

 

 

 

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YDSE with thin slab

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