JEE Mains · Physics · STD 12 - 10. Wave optics
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)
- A \(\frac{\lambda }{{2\,\left( {\mu - 1} \right)}}\)
- B \(\frac{\lambda }{{\,\left( {\mu - 1} \right)}}\)
- C \(\frac{\lambda }{{\,\left( {2\mu - 1} \right)}}\)
- D \(\frac{{2\lambda }}{{\,\left( {\mu - 1} \right)}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\lambda }{{\,\left( {\mu - 1} \right)}}\)
Step-by-step Solution
Detailed explanation
\(\Delta x=(\mu-1)\, t=1 \lambda\) for one maximum shift \(t=\frac{\lambda}{\mu-1}\)
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