Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. Part AIn terms of the intensity I0 at the peak of the central maximum, what is the intensity of the light at the screen at the distance 1.00 mm from the center of the central maximum?

The intensity of light from the 1mm from the central maximu is  

Explanation:

From the question we are told that

                         The wavelength is

                         The width of the slit is  

                          The distance from the screen is  

                           The intensity at the central maximum is

                          The distance from the central maximum is

        Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

    Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

if the angle between the the light at z and the central maximum is small

Then  

   Which implies that

substituting this into the equation for the intensity

given that

   We have that

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