Question

White coherent light (400 nm - 700 nm) is sent through the slits of a young's double slit experiment (figure 17-E3). The separation between the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (aong 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 wavelength (s) willl have a strong intensity?

Answer 1

Given That, $\lambda =\left(400nmto700nm\right),$
$d=0.5mm=0.5\times {10}^{-8}m,$
$D=50cm=0.5m$
and on the screen $y_n$ = 1 mm $=1\times {10}^{-3}$ m
(a) We know that for zero intensity (dark fringe)
$y_n=\frac{(2n+1)}{2{\lambda }_n}\frac{D}{d}Wheren=0,1,2,\cdots \Rightarrow {\lambda }_n=\frac{2}{(2n+1)}\frac{y_n}{D}=\frac{2}{(2n+1)}\times \frac{{10}^{-3}\times \left(0.05\right)\times {10}^{-3}}{\left(0.5\right)}=\frac{2}{(2n+1)}\times {10}^{-6}m=\frac{2}{(2n+1)}\times {10}^{3}nmifn=1,{\lambda }_1=\left(\frac{2}{3}\right)\times 1000=667nm.Ifn=1,{\lambda }_2=\left(\frac{2}{5}\right)\times 1000=400nm$
So, the light waves of wavelength 400 nm and 667 nm will be absent from the outcming light.
(b) For strong intensity (bright fringes) at the hold,
$Y_n=n{\lambda }_n\frac{D}{d}$
${\lambda }_n=y_n\frac{d}{nD}When,n=1,lambd{a}_1=y_n\frac{d}{D}={10}^{-3}\times \left(0.5\right)\times \frac{0^{-3}}{0.5}={10}^{-3}m=100nm.$
1000 nm is not present in the range 400 nm- 700 nm
Again , when n = 2,
${\lambda }_2=y_n\frac{d}{2D}=500nm$
So, the wavelength which will have strong intensity is 500 nm.