JEE Mains · Physics · STD 12 - 10. Wave optics
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 gets shifted by a distance equal to \(n\) fringe widths. If the wavelength of light used is \(\lambda ,\,t\) will be
- A \(\frac{{2nD\lambda }}{{a\left( {\mu - 1} \right)}}\)
- B \(\frac{{nD\lambda }}{{a\left( {\mu - 1} \right)}}\)
- C \(\frac{{2D\lambda }}{{a\left( {\mu - 1} \right)}}\)
- D \(\frac{{n\lambda }}{{\left( {\mu - 1} \right)}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{n\lambda }}{{\left( {\mu - 1} \right)}}\)
Step-by-step Solution
Detailed explanation
Path difference at central maxima \(\Delta x=(\mu-1)\, t,\) whole pattern will shift by same amount which will be given by \((\mu-1) \,t \frac{D}{d}=n \frac{\lambda D}{d},\) according to eh question \(t=\frac{n \lambda}{(\mu-1)}\) No option is matching, therefore question should…
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