Question:
In a double slit experiment, when a thin film of thickness t having refractive index \(\mu\). 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 (\(\lambda\) is the wavelength of the light used) :
In a double slit experiment, when a thin film of thickness t having refractive index \(\mu\). 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 (\(\lambda\) is the wavelength of the light used) :
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A film adds extra optical path \((\mu-1)t\); a shift of one fringe means this path equals one wavelength.
Updated On: Nov 5, 2026
- \(\dfrac{\lambda}{2(\mu-1)}\)
- \(\dfrac{\lambda}{(2\mu-1)}\)
- \(\dfrac{2\lambda}{(\mu-1)}\)
- \(\dfrac{\lambda}{(\mu-1)}\)
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The Correct Option is D
Solution and Explanation
Fringe shift due to a film: \(\Delta x = \dfrac{(\mu-1)tD}{d}\). Fringe width: \(\beta = \dfrac{\lambda D}{d}\).
Shift of one fringe width: \(\Delta x = \beta\), so \(\dfrac{(\mu-1)tD}{d} = \dfrac{\lambda D}{d}\).
Hence \(t = \dfrac{\lambda}{\mu-1}\).
Answer: (4) \(\dfrac{\lambda}{(\mu-1)}\)
Shift of one fringe width: \(\Delta x = \beta\), so \(\dfrac{(\mu-1)tD}{d} = \dfrac{\lambda D}{d}\).
Hence \(t = \dfrac{\lambda}{\mu-1}\).
Answer: (4) \(\dfrac{\lambda}{(\mu-1)}\)
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