Question

In Young's experiment, 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 $\lambda$,will be

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

Answer 1

The correct option is C $\frac{d}{81D\lambda }$

We know,

Fringe width, $\beta =\frac{{\lambda }^{\prime }{D}^{\prime }}{d^{\prime }}=\frac{\left(3\lambda \right)\left(3D\right)}{\frac{d}{3}}=27\frac{\lambda D}{d}$

Number of fringes $=\frac{\frac{1}{3}}{\beta }=\frac{d}{\left(3\right)\left(27\lambda D\right)}=\frac{d}{81D\lambda }$