Optics Ques 129
- In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$ the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is
(a) $(2 n+1) \frac{\lambda}{2}$
(b) $(2 n+1) \frac{\lambda}{4}$
(c) $(2 n+1) \frac{\lambda}{8}$
(d) $(2 n+1) \frac{\lambda}{16}$
(2013 Adv.)
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Answer:
Correct Answer: 129.(b)
Solution:
Formula:
- Given,
$ \begin{aligned} & I=I _{\max } \cos ^{2} \frac{\phi}{2} \quad …….(i) \\ & I=\frac{I _{\max }}{2} \quad …….(ii) \end{aligned} $
$\therefore$ From Eqs. (i) and (ii), we have
$ \phi=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2} $
Or path difference, $\Delta x=(\frac{\lambda}{2 \pi} )\cdot \phi$
$ \therefore \quad \Delta x=\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4} . .,( \frac{2 n+1}{4} )\lambda $
BETA
