Question

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?

Answer 1

Answer:

The intensity of light from the 1mm from the central maximu is  $I = 0.822I_o$

Explanation:

From the question we are told that

                         The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

                         The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$  

                          The distance from the screen is  $D = 3.00m$

                           The intensity at the central maximum is $I_o$

                          The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

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

Then we can represent this intensity as

                     $I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

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

            $sin \theta = \frac{z}{D}$

           

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

Then  $sin \theta = \theta$

   Which implies that

              $\theta = \frac{z}{D}$

substituting this into the equation for the intensity

             $I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

   We have that

              $I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$

                 $=I_o [\frac{sin(0.760)}{0.760}] ^2$

                 $I = 0.822I_o$