Question

A screen containing two slits 0.100mm apart is 1.50m from a viewing screen. Light of wavelength 633nm falls on the slits. Approximately how far apart are the bright fringes on the viewing screen?

Answer 1

distance = 9.495 mm

Explanation:

Given data

slits distance = 0.100 mm

viewing screen distance = 1.50 m

wavelength = 633 nm

to find out

how far apart are the bright fringes on the viewing screen

solution

we use here double slit diffraction condition that is

d sin(θ) = m λ .......................1

here m = 0, 1 , 2 , 3 .....

and here for θ small so

sin(θ) = tan(θ) =
`(x)/(l)`

so x = l sin (θ)

so from equation 1

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

and

x1 =
`(\lambda l)/(d)` ...............2

x2 =
`(2\lambda l)/(d)` ..............3

so distance x2 - x1 will be

distance =
`(\lambda l)/(d)`

distance =
`(633*10^(-9)*1.5)/(0.1*10^(-3))`

distance = 9.495 mm