Question

In a Young’s double slit experiment two rays of monochromatic light emerge from the slits and meet at a point on a distant screen.The point on the screen where these two rays meet is the eighth order bright fringe...?

The
difference in the distances that the two rays travel is `4.57 * 10^(-6)` m.
What is the wavelength of the light? Find the position of the second
order maximum if the distance between the slits is d= 2*10m and the
slits-screen distance is L=3m ?

Answer 1

`lambda=5.71*10^(-7)color(white)(l)"m"`
`s_2=1.71*10^(-7)color(white)(l)"m"`

Explanation:

The forumla

`Delta lambda_n=n*lambda`

gives the two monochromatic (of wavelength `lambda`) beam's path difference required to generate the `n`-th maximum interference. For example, `Delta lambda=0*lambda=0` for `n=0` gives the path difference required to generate the central maximum whereas letting `n=8` shall give that of the eighth maximum.

The question states that `Delta lambda=4.57*10^(-6) color(white)(l)"m"` so we would expect

`Delta lambda_8=8*lambda=4.57*10^(-6) color(white)(l)"m"`

hence `lambda=(Delta lambda_8)/8=5.71*10^(-7)color(white)(l)"m"`

`n=2` for the second maximum. Therefore there shall be a path difference of
`Delta lambda_2=2*lambda=1.14*10^(-6)color(white)(l)"m"`

Referring to the diagram,
given that `lambda` is considerably smaller than `d`,
we can assume that `(Delta lambda_2)/(d)=sin(theta)~~tan(theta)=s_2/D`

where `s_2` the distance between the central and the second maximum `d` the separation of the two slits, and `D` the distance between the slits and the screen.

Solving for `s_2` gives

`s_2=(Delta lambda_2)/(d)*D=1.14*10^(-6)color(white)(l)"m"`

Note that the value of `d` given in the in the question seems to be off by a couple of magnitudes. See if you're able to find `s_2` with a smaller value of `d`.