Question

Two sources of light of wavelength 686 nm are separated by a horizontal distance x. They are 3 m from a vertical slit of width 0.5 mm. What is the least value of x for which the diffraction pattern of the sources can be resolved by Rayleigh's criterion?

Answer 1

X = 5.48 mm

Step-by-step explanation:

for single slit

By Rayleigh criterian

`sin\theta _R = (\lambda)/(d)`

`sin\theta _R = tan\theta _R ≈ (x)/(3)`

where d = slit width =0.5 mm

wavelength
`\lambda = 686 nm`

`(x)/(3) =(686*10^(-9))/(0.5*10^(-3))`

therefore maximum of value of X can be calculated from above

`X = 5.48*10^(-3) m`

X = 5.48 mm