Question

In Young's double slit experiment the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe and 3rd dark fringe from the centre are 9mm apart. What is the wavelength of light used?

• A. 2000A
  
• B. 4000A
  
• C. 6000A
  
• D. 8000A

Answer 1

The correct option is C

6000A

The distance of the $m^{th}$ bright fringe from the central fringe is
$y_m=\frac{m\lambda D}{d}=m\beta where\beta =\lambda D/disthefringewidth.\therefore y_9=9\beta \left(i\right)$
The distance of the mth dark fringe from the central fringe is
$y_m^{\prime }=(m-\frac{1}{2})\frac{\lambda D}{d}=(m-\frac{1}{2})\beta \therefore y_2^{\prime }=\frac{3}{2}\beta \left(ii\right)$
From Eqs. (i) and (ii), we get $y_9-y_2^{\prime }=9\beta -\frac{3}{2}\beta =\frac{15}{2}\beta$
It is given that $y_9-y_2^{\prime }=9.0mm.Hence\beta =\frac{9.0\times 2}{15}=1.2mmNow\lambda =\beta d/D.Substitutingfor\beta ,dandD,weget\lambda =6\times {10}^{-7}m=6000A$