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Wave Optics question

2019 · 9 Apr · Shift 1 · Q56
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Wave Optics question

2019 · 9 Apr · Shift 1 · Q56

JEE MainPhysicsWave OpticsMCQ+4 / −1
The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index μ is put in front of one of the slits, the central maximum gest shifted by a distance equal to n fringe widths. If the wavelength of light used is λ\lambdaλ, t will be : JEE Main 2019 (Online) 9th April Morning Slot Physics - Wave Optics Question 117 English
  1. A
    Dλa(μ−1){{D\lambda } \over {a\left( {\mu - 1} \right)}}a(μ−1)Dλ​
  2. B
    2nDλa(μ−1){{2nD\lambda } \over {a\left( {\mu - 1} \right)}}a(μ−1)2nDλ​
  3. C
    2Dλa(μ−1){{2D\lambda } \over {a\left( {\mu - 1} \right)}}a(μ−1)2Dλ​
  4. D
    nλ(μ−1){{n\lambda } \over {\left( {\mu - 1} \right)}}(μ−1)nλ​
View written solutionFree

Correct answer: D

  1. Effect of inserting a thin sheet in YDSE

When a thin transparent sheet of thickness ttt and refractive index μ\muμ is introduced in front of one slit, the optical path in that arm increases by

Δ=(μ−1)t\Delta = (\mu-1)tΔ=(μ−1)t

This produces a shift of the entire fringe pattern.

  1. Relation between path difference and fringe shift

In Young’s double slit experiment, fringe width is

β=Dλa\beta = \frac{D\lambda}{a}β=aDλ​

If the central maximum shifts by a distance equal to nnn fringe widths, then the shift is

x0=nβx_0 = n\betax0​=nβ

Also, shift due to introduction of sheet is

x0=Da(μ−1)tx_0 = \frac{D}{a}(\mu-1)tx0​=aD​(μ−1)t
  1. Equate the two expressions for shift
Da(μ−1)t=nβ=nDλa\frac{D}{a}(\mu-1)t = n\beta = n\frac{D\lambda}{a}aD​(μ−1)t=nβ=naDλ​

Cancel Da\frac{D}{a}aD​ from both sides:

(μ−1)t=nλ(\mu-1)t = n\lambda(μ−1)t=nλ

Hence,

t=nλμ−1t = \frac{n\lambda}{\mu-1}t=μ−1nλ​
  1. Match with options

This corresponds to:

nλμ−1\boxed{\frac{n\lambda}{\mu-1}}μ−1nλ​​

So, the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They agree.

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