Question:
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 \(\mu\) 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 \(\lambda\), \(t\) will be :
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 \(\mu\) 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 \(\lambda\), \(t\) will be :
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The sheet adds an extra optical path \((\mu-1)t\); the central maximum shifts by \(n\) fringes when this equals \(n\lambda\).
Updated On: Nov 5, 2026
- \(\frac{D\lambda}{a(\mu-1)}\)
- \(\frac{2nD\lambda}{a(\mu-1)}\)
- \(\frac{2D\lambda}{a(\mu-1)}\)
- \(\frac{n\lambda}{(\mu-1)}\)
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The Correct Option is D
Solution and Explanation
A sheet of thickness \(t\) and index \(\mu\) adds an optical path difference \((\mu-1)t\).
Shift of central maximum: \(y_0=\frac{D}{a}(\mu-1)t\).
Fringe width: \(\beta=\frac{\lambda D}{a}\). Given \(y_0=n\beta\):
\(\frac{D}{a}(\mu-1)t=\frac{n\lambda D}{a}\Rightarrow t=\frac{n\lambda}{\mu-1}\).
Answer: (4) \(\frac{n\lambda}{(\mu-1)}\)
Shift of central maximum: \(y_0=\frac{D}{a}(\mu-1)t\).
Fringe width: \(\beta=\frac{\lambda D}{a}\). Given \(y_0=n\beta\):
\(\frac{D}{a}(\mu-1)t=\frac{n\lambda D}{a}\Rightarrow t=\frac{n\lambda}{\mu-1}\).
Answer: (4) \(\frac{n\lambda}{(\mu-1)}\)
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