Optics Ques 137
- In the ideal double-slit experiment, when a glass-plate (refractive index $1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$ ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is
(2002, 2M)
(a) $2 \lambda$
(b) $\frac{2 \lambda}{3}$
(c) $\frac{\lambda}{3}$
(d) $\lambda$
Show Answer
Answer:
Correct Answer: 137.(a)
Solution:
Formula:
- Path difference due to slab should be integral multiple of $\lambda$ or
$ \Delta x=n \lambda \quad \text { or } \quad(\mu-1) t=n \lambda \quad n=1,2,3, \ldots $
or $\quad t=\frac{n \lambda}{\mu-1}$
For minimum value of $t, n=1$
$ \therefore \quad t=\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}=2 \lambda $
BETA
