Question

Light of wavelength 550 nm falls on a

slitthat is 3.00x10^-3mm wide. How far from the central maximumwill
the first diffraction maximum fringe be if the screen is
10.0maway?

Answer 1

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Step-by-step explanation:

We can describe single slit diffraction phenomenon with the equation:

`a\sin\theta=m\lambda` (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum (
`y_(m)`) are small compared to the distance between the screen and the slit (x), we can approximate
`\sin\theta\approx\tan\theta=(y)/(x)`, using this on (1):

`a(y_(m))/(x)=m\lambda`

solving for y

`y_(m)= (mx\lambda)/(a)`

Note that
`y_(m)`is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find
`y_(1)`,
`y_(2)` add them and divide by two:

`y_(1)= ((1)(10.0m)(550*10^(-9)\,m))/(3.00*10^(-6)\,m)`

`y_(1)= 1.8 m`

`y_(2)= ((2)(10.0m)(550*10^(-9)\,m))/(3.00*10^(-6)\,m)`

`y_(1)= 3.6 m`

`maximum = (1.8+3.6)/(2)=2.7m`