AP EAMCET · PHYSICS · Wave Optics
In young's double slit experiment, light of wavelength \(480 \mathrm{~nm}\) is incident on two slits separated by a distance of \(4 \times 10^{-4} \mathrm{~m}\). If a thin plate of thickness \(1.4 \times 10^{-6} \mathrm{~m}\) and refractive index \(\frac{13}{7}\) is placed between one of the slits and screen, the phase difference introduced at the position of central maxima is
- A \(5 \pi\)
- B \(\frac{7}{3} \pi\)
- C \(\frac{7}{4} \pi\)
- D \(4 \pi\)
Answer & Solution
Correct Answer
(A) \(5 \pi\)
Step-by-step Solution
Detailed explanation
The YDSE setup is shown below Additional path difference introduced by medium plate, \(\Delta L=(\mu-1) t\) As, a path difference of one wavelength is equals to a phase difference \(\Delta \phi=\frac{\Delta L \times 2 \pi}{\lambda}\) so,…
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