Question

In a Youngs double experiment, the slits are $1.5mm$ apart. When the slits are illuminated by a monochromatic light source and the screen is kept $1m$ apart from the slits, width of $10$ fringes is measured as $3.93mm$. Calculate the wavelength of light used. What will be the width of $10$ fringes when the distance between the slits and the screen is increased by $0.5m$. The source of light used remains the same.

Answer 1

We know,

Fringe width=$\frac{\lambda D}{d}$

Here fringe width in 1st case =$\frac{3.93}{10}=0.393mm$

$D=1000mm$, $d=1.5mm$

$\lambda =\frac{0.393\times 1.5}{1000}=\frac{3.393\times 1.5}{10000}$

$\lambda =5.0895\times {10}^{-4}mm$

$=508.95nm$

Now , fringe width of 10 fringes when $d=1.5+0.5=2.0m$

Let initial and final fringe width of 10 fringes be $f_i,f_f$

$\frac{f_i}{f_f}=\frac{d_f}{d_i}=\frac{2}{1.5}=\frac{4}{3}$

$f_f=\frac{3.93\times 4}{3}=52.4mm$