Verified Solution • 2025–26 Session moderate

QJEEM19Ph571a: JEE Main 2019 Physics - Class 12 JEEM19Ph571a Solution: In a double slit experiment, when a thin film of thickness t…

In a double slit experiment, when a thin film of thickness t having refractive index μ\mu . is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( λ\lambda is the wavelength of the light used) :

Multiple Choice Options

  • A λ2(μ−1){\lambda \over {2\left( {\mu - 1} \right)}}
  • B λ(2μ−1){\lambda \over {\left( {2\mu - 1} \right)}}
  • C 2λ(μ−1){{2\lambda } \over {\left( {\mu - 1} \right)}}
  • D λ(μ−1){\lambda \over {\left( {\mu - 1} \right)}}

Step-by-Step Verified Solution

As we know, Path difference introduced by thin film,

Δ=(μ−1)t\Delta=(\mu-1) t .......(i)

JEE Main 2019 (Online) 12th April Morning Slot Physics - Wave Optics Question 103 English Explanation
and if fringe pattern shifts by one frings width, then path difference,

Δ=1×λ=λ\Delta=1 \times \lambda=\lambda .......(ii)

So, from Eqs. (i) and (ii), we get

(μ−1)t=λ(\mu-1) t=\lambda

⇒\Rightarrow t=λμ−1t=\frac{\lambda}{\mu-1}

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