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 7.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m=0 and m=1 maxima to be 35 cm. What slit separation is required in order to produce the desired interference pattern?

Answer 1

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m