Question

In Youngs expt., the distance between two slits is d/3 and the distance between the screen and the slits is 3D. The number of fringes in 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

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

Triangle width = $\frac{\lambda D}{d}$
= $\frac{3\lambda \left(3D\right)}{\frac{d}{3}}$
= $\frac{27\lambda D}{d}$
no. of fringes = $\frac{1}{3}\times \frac{27\lambda D}{d}$
= $\frac{d}{81\lambda D}$