White coherent light `(400 nm-700 nm)` is sent through the slits of a YDSE. `D=0.5 mm`, D=50 cm. There is a hole in the screen at a point `1.0 mm` away (along the width of the fringes) from the central line.
(a) Which wavelength will be absent in the light coming from the hole?
(b) Which wavelength(s) will have a strong intensity?

Answer

A, B, D

Explanation

Given that `lamda=(400nm to 700nm).,
`d=0.5mm=0.5xx10^-3m`
D=50cm=0.5m`
and on the screen `y_n=1mm=1xx10^-3m`
a. We know that for zero intensity (dark fringe)
`y_n=((2n+1))/(2lamda_n) D/d`
where n=0,1,2,……..
`rarr lamda_n=2/((2n+1))(y_nd)/D`
`=2/((2n+1))xx((10^-3xx()0.05)xx10^-3)/((0.5))`
`=2/((2n+1)xx10^-6m`
`=2/((2n+1))xx10^3nm`
If `n=1`
`lamda_1=(2/3)xx1000`
`=667nm`
`If n=1`
`lamda_2=(2/5)xx1000`
`=400nm`
So the light waves of wavelength 400 nm and 667 nm will be absent from the outcoming light.
b. For strong intensity (bright fringes) at the hold
`y_n=nlamda_nD/d`
`lamda_n=y_n d/(nD)`
`when n=1, lamda_1=y_n d/D`
`=10^-3xx(0.5)xx10^-3/0.5`
`=10^-3m=100nm`
1000 nm is not present in the range 400 nm -700 nm
Again when n=2
`lamda_2=y_n=y_n d/(2D)=500nm`
So the wavelength which will have strong intensity is 500 nm.