Question

The slits in Young's double slit experiment, are $0.5mm$ apart and interference pattern is observed on a screen distant $100cm$ from the slits. It is found that the $9th$ bright fringes is at a distance of $8.835mm$ from the second dark fringe. The wavelength of light will be

• A. $7529A^o$
• B. $6253A^o$
• C. $6779A^o$
• D. $5890A^o$

Answer 1

Answer: (D)

$5890A^o$

Let $y_1,y_2$ be the positions of the 2nd dark fringe and 9th bright fringe respectively.

Therefore, $\frac{y_{1}d}{D}=\frac{3\lambda }{2}$

And $\frac{y_{2}d}{D}=9\lambda$

Subtracting the above equations and rearranging gives,

$\lambda =\frac{2(y_2-y_1)d}{15D}=5890A^{\circ }$