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

Answer:

X = 5.48 mm

Explanation:

for single slit

By Rayleigh criterian

$sin\theta _R = \frac{\lambda}{d}$

$sin\theta _R = tan\theta _R ≈ \frac{x}{3}$

where d = slit width =0.5 mm

wavelength $\lambda = 686 nm$

$\frac{x}{3} =\frac{686*10^{-9}}{0.5*10^{-3}}$

therefore maximum of value of X can be  calculated from above

$X = 5.48*10^{-3} m$

X = 5.48 mm