White coherent light `(400 nm-700 nm)` is sent through the slits of a Young's double slit experiment (as shown in the figure). The separation between the slits is `1 mm` and the screen is `100 cm` away from the slits. There is a hole in the screen at a point `1.5 mm` away (along the width of the fringes) from the central line. `(a)` For which wavelength `(s)` there will be minima at that point ? `(b)` which wavelength `(s)` will have a maximum intensity?

Answer

(a) `428 nm, 600 nm,
(b)`500 nm`

Explanation

As `lambda lt lt d, rArr beta=(lambdaD)/(d)` can be used
For minima
(a) `y=(n+(1)/(2))beta.=(n+(1)/(2))(lambdaD)/(d)`
`rArr lambda=(1.5xx10^(-3)xx1xx10^(-3))/(1xx(n+(1)/(2))) =(1.5xx10^(-6))/(n+(1)/(2))`
`n=1, lambda=(2)/(3)xx1.5xx10^(-6)=1000 nm`
`n=2, lambda=(2)/(5)xx1.5xx10^(-6)= 600 nm`
`n=3, lambda=(2)/(7)xx1.5xx10^(-6)=428 nm`
Putting integral values of `'n'`
`n=1 , lambda=1000 nm`
`n=2, lambda=600 nm`.
`n=3, lambda=428 nm`
So only `lambda=428 nm and lambdad=6000 nm,` will have minima at the hole. Hence they will be absent in the light coming out.
(b) `1.5 mm = n beta`.
`1.5 mm = n((lambdaD)/(d))`
`rArr lambda=(1.5xx10^(-3)xx1xx10^(-3))/(nxx100xx10^(2))`
for `n=1. lambda = 1500 nm`.
for `n =2. lambda = 750 nm`.
for `n=3. lambda=500 nm`.
Hence only `lambda=500 nm` will have maximum intensity.