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 $1\mathrm{cm}$across, and you estimate that the distance from the window shade to the wall is about $3\mathrm{m}.$. Estimate (a) the average wavelength of the sunlight (in $\mathrm{nm}$ ) and (b) the diameter of the pinhole (in $\mathrm{mm}$ ).

Short answer

a) The average wavelength of the sunlight$550\mathrm{nm}$

b) The diameter of the pinhole$0.4\mathrm{mm}$

Step-by-step solution

Average Wavelength

vaguely remember that when using λ and aperture diameter D, the expression for the wavelength of the beams emanating from a completely circular probe at position L from the pupil is

$w=\frac{2.44\times \lambda L}{D}$

This means that the beam diameter is linearly proportional to the wavelength used. Therefore, if we want to consider an average wavelength used, assuming the intensity per wavelength was constant, this average will be just the simplest arithmetic mean; in our case, half the sum of the lowermost and uppermost wavelengths of the visible spectrum . That is,

$$\begin{gathered} {\lambda }_{\mathrm{ovg}}=\frac{{\lambda }_{min}+{\lambda }_{max}}{2} \\ =\frac{400+700}{2} \\ =550\mathrm{nm} \end{gathered}$$

Diameter

They can already obtain the aperture's length as

$D=\frac{2.44\times \lambda L}{w}$

$$\begin{gathered} D=\frac{2.44\times 5.5\times {10}^{-7}\times 3m}{0.01} \\ =4.03\times {10}^{-4}\mathrm{m} \\ =0.4\mathrm{mm} \end{gathered}$$(Numerically)