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 5.0 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

The distance of separation is
`d = 9.04 *10^(-6 ) \ m`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 633\ nm = 633 *10^(-9) \ m`

The distance of the screen is
`D = 5.0 \ m`

The distance between the fringes is
`y = 35 \ cm = 0.35 \ m`

Generally the distance between the fringes is mathematically represented as

`y = ( \lambda * D )/(d )`

Here d is the distance of separation between the slit

=>
`d = ( \lambda * D )/(y )`

=>
`d = ( 633 *10^(-9) * 5 )/( 0.35 )`

=>
`d = 9.04 *10^(-6 ) \ m`