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

Answer:

The distance of separation is $d = 9.04 *10^{-6 } \ m$

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

       $y = \frac{ \lambda * D }{d }$

Here d is the distance of separation between the slit

=>    $d = \frac{ \lambda * D }{y }$

=>    $d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }$

=>   $d = 9.04 *10^{-6 } \ m$