Question

In Young's expt., the distance between two slits is $\frac{d}{3}$ and the distance between the screen and the slits is 3 D. The number of fringes in $\frac{1}{3}$ m on the screen, formed by monochromatic light of wavelength 3$\lambda$, will be

• A. $\frac{d}{9D\lambda }$
• B. $\frac{d}{27D\lambda }$
• C. $\frac{d}{81D\lambda }$
• D. $\frac{d}{D\lambda }$

Answer 1

Answer: (C)

$\frac{d}{81D\lambda }$

$\beta =\frac{{\lambda }^{{}^{\prime }}{D}^{{}^{\prime }}}{d{}^{\prime }}=\frac{3\lambda 3D}{d/3}=27\frac{\lambda D}{d}$ .
No.of fringes = $\frac{1/3}{\beta }$ = $\frac{d}{81\lambda D}$