Question

Light of wavelength 710 nm passes through two narrow slits 0.66 mm apart. The screen is 2.00 m away. A second source of unknown wavelength produces its second-order fringe 1.25 mm closer to the central maximum than the 710-nm light. What is the wavelength of unknown light?

Answer 1

The wavelength is 503 nm

Step-by-step explanation:

Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light

i.e dsinθ = m
`\lambda`

Where
`\lambda` is the wavelength of light and m is the order of the fringe

Looking at θ to be very small , sin θ can be approximated to θ

and
`\theta \approx (x)/(l)`

Substituting this into the above equation

`d[(x)/(l) ] =m\lambda`

making x the subject

`x =(m\lambda l)/(d)`

This above equation will give the value of the distance of the
`m^(th)` order fringe of the wavelength
`\lambda` from the central fringe

Replacing with the value given in the question we have

`\lambda` = 710 nm m = 2 d =0.66 mm , l = 2.0 m

`x = ((2)(710nm)(2.0m)[(10^9)/(1m) ])/((0.66mm)((10^6)/(1mm) ))`

`=(4.303*10^6nm)[((1)/(10^6)mm )/(1nm) ]`

`=4.303mm`

The separation of the second fringe from central maximum is 4,303 mm

To obtain the separation of the second order fringe of the unknown light from central maximum

`x' = 4.303mm - 1.25 mm = 3.053mm`

Now to obtain the wavelength of this second source

from
`x = (m\lambda l)/(d)`

`\lambda' = (x'd)/(ml)`

Now substituting 3,053 mm for
`x'` 2.0 mm for l , 0.66 mm for d and 2 for m in the above formula

`\lambda' =((3.053mm)(0.66mm))/((2)(2.0)((10^3mm)/(1m) ))`

`= (503.7*10^(-6)mm)((10^6nm)/(1mm) )`

`=503.7nm`