Optics 6 Question 21

23. 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

Solution:

  1. 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 $t, n=1$

$$ \therefore \quad t=\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}=2 \lambda $$



NCERT Chapter Video Solution

Dual Pane