Question

In the ideal double-slit experiment , when a glass-plate (refractive index $\mu =1.5$) of thickness $t$ is introduced in the path of one of the interfering beams having wavelength $\left(\lambda \right),$ the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is-

• A. $2\lambda$
• B. $\frac{2\lambda }{3}$
• C. $\frac{\lambda }{3}$
• D. $\lambda$

Answer 1

The correct option is A $2\lambda$

Path difference due to slab should be integral multiple of $\lambda$ i.e.,

$\Delta x=n\lambda$

$\Rightarrow (\mu -1)t=n\lambda [n=1,2,3,..]$

$\Rightarrow t=\frac{n\lambda }{\mu -1}$

For minimum value of $t,n=1$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.