Question

White coherent light $(400nm-700nm)$ is sent through the slits of a Youngs double slit experiment . The separation between the slits is 0.5 mm and the screen is $50cm$ away from the slits. There is a hole in the screen at a point $1.0mm$ away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole ? (b) which will have a strong intensity ?

Answer 1

Given:

The wavelength of the light is $\lambda =\left(400nmto700nm\right)$

The separation between the slits is $d=0.5mm=0.5\times {10}^{-3}m$,

The distance between the slits and screen is $D=50cm=0.5m$

and on the screen $y_n=1mm=1\times {10}^{-3}m$

a) We know that for zero intensity (dark fringe)

$y_n=\left(\frac{2n+1}{2}\right)\frac{{\lambda }_{n}D}{d}$ where $n=0,1,2,.....$

$\Rightarrow {\lambda }_n=\frac{2}{(2n+1)}\frac{{\lambda }_{n}d}{D}$

$=\frac{2}{2n+1}\times \frac{{10}^{-3}\times 0.5\times {10}^{-3}}{0.5}$

$\Rightarrow \frac{2}{(2n+1)}\times {10}^{-6}=\frac{2}{(2n+1)}\times {10}^{3}nm$

If $n=1,{\lambda }_1=\left(2/3\right)\times 1000=667nm$

If $n=1,{\lambda }_2=\left(2/5\right)\times 1000=400nm$

So, the light waves of wavelengths $400nm$ and $667nm$ will be absent from the out coming light.

b) For strong intensity (bright fringes) at the hole

$y_n=\frac{n{\lambda }_{n}D}{d}\Rightarrow {\lambda }_n=\frac{y_{n}d}{nD}$

When $n=1$

${\lambda }_1=\frac{y_{n}d}{D}=\frac{{10}^{-3}\times 0.5\times {10}^{-3}}{0.5}={10}^{-6}m=1000nm$

$1000nm$ is not present in the range $400nm-700nm$

Again, where $n=2,{\lambda }_2=\frac{y_{n}d}{2D}=500nm$

So, the only wavelength which have strong intensity is $500nm$.