Question

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m

Answer 1

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

$d (\frac{y}{L}) =$ mλ

$d (\frac{y}{L}) = (1)(633nm)$

$d(\frac{0.25}{4} ) = (1)(633nm)$

d = 1.0128×10⁻⁵m