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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?

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Answer:

The distance of separation is [tex]d = 9.04 *10^{-6 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is [tex]\lambda = 633\ nm = 633 *10^{-9} \ m[/tex]

     The  distance of the screen is [tex]D = 5.0 \ m[/tex]

     The  distance between the fringes is  [tex]y = 35 \ cm = 0.35 \ m[/tex]

       

Generally the distance between the fringes is mathematically represented as

       [tex]y = \frac{ \lambda * D }{d }[/tex]

Here d is the distance of separation between the slit

=>    [tex]d = \frac{ \lambda * D }{y }[/tex]

=>    [tex]d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }[/tex]

=>   [tex]d = 9.04 *10^{-6 } \ m[/tex]

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