Question

Young's experiment is performed with light of wavelength 502nm from excited helium atoms. Fringes are measured carefully on a screen 1.35m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.4mm from the center of the central bright fringe.
What is the separation of the two slits?

Answer 1

Answer:

d = 1.30 mm

Explanation:

given,

wavelength of the light source (λ)= 502 nm

slits is separated (d) = ?

distance to form interference pattern(D) = 1.35 m  

20 th fringe

y = 10.4 mm = 0.0104 m

now,separation between to slits

 $d = \dfrac{m\lambda\ D}{y}$

where y is the fringe width

 $d = \dfrac{20 \times 502\times 10^{-9}\ \times 1.35}{0.0104}$

 $d = \dfrac{13354\times 10^{-9}}{0.0104}$

      d = 1.30 mm

separation between to slits is equal to d = 1.30 mm