Question

In young double slit experiment, while using a source of light of wavelength 5000 Å, the fringe width 

obtained is 0.6 cm. If the distance between the screen and the slit is reduced to half, what should be the 

 wavelength of the source to get fringes of 0.3cm wide?

Answer 1

$\Delta y_1=\frac{L}{d}\lambda_1$

$\Delta y_2=\frac{L}{2d}\lambda_2$

So, we have

$\Delta y_1/\Delta y_2=\frac{L}{d}\lambda_1/\frac{L}{2d}\lambda_2\to 2\lambda_1/\lambda_2\to$

$\lambda_2=2\lambda_1\Delta y_2/\Delta y_1=2\cdot5000\cdot0.3/0.6=5000\ Å$ . Answer