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Harman [31]
3 years ago
7

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s

mall 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.
Physics
1 answer:
gayaneshka [121]3 years ago
4 0

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

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