Question

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

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

Answer 1

The correct option is D 5890 $A^o$

$9^{th}brightfringe=\frac{9\lambda D}{d}$
$2^{nd}darkfringe=\frac{3}{2}\frac{\lambda D}{d}$
So, $\frac{9\lambda D}{d}-\frac{3}{2}\frac{\lambda D}{d}=8.835\times {10}^{-3}m$
$\frac{15}{2}\frac{\lambda D}{d}=8.835\times {10}^{-3}$
$\lambda =\frac{8.835\times {10}^{-3}\times 2\times 0.5\times {10}^{-3}}{15\times 1}$
$=0.589\times {10}^{-6}$
$=5890\times {10}^{-10}m$
$=5890\dot{A}$