Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cmcm across, and you estimate that the distance from the window shade to the wall is about 5 mm.
Required:
Estimate the diameter of the pinhole.

Answer 1

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.  

Answer:

The diameter is  $d =0.000336 m$

Explanation:

     From the question we are told that

            The central maxima is $D= 2cm = \frac{2}{100} = 0.02m$

            The distance from the window shade is $L = 5m$

     The  average wavelength of the  sun is mathematically evaluated as

                         $\lambda_{ave } = \frac{\lambda_i + \lambda_f}{2}$

 Generally the visible light spectrum  has a wavelength  range  between  400 nm  to 700 nm  

        So  the initial wavelength of the sun is $\lambda _i = 400nm$

           and the final wavelength is  $\lambda_f = 700nm$

  Substituting this into the above equation

                 $\lambda_{sun} = \frac{400nm +700nm}{2}$

                        $= 550nm$

The diameter is evaluated as

              $d = \frac{2.44 \lambda_{sun} L}{D}$

substituting values

              $d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}$

                $d =0.000336 m$