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

Respuesta :

Answer:

a)[tex]\lambda _{avg}=550 nm[/tex]

b)D=0.402 mm

Explanation:

We know that visible range of sun light is 400 nm to 700 nm.

So the average wavelength

[tex]\lambda _{avg}=\dfrac{\lambda _1+\lambda _2}{2}[/tex]

[tex]\lambda _{avg}=\dfrac{400+700}{2}[/tex]

[tex]\lambda _{avg}=550 nm[/tex]

To find the diameter

We know that the relation

[tex]D=\dfrac{2.44\lambda L}{w}[/tex]

Where D=Diameter of pin hole

          λ=Average wavelength

           w=Central distance

        L=length

So now by putting the values

[tex]D=\dfrac{2.44\times 550\times 10^{-9}\times 3}{0.01}[/tex]

D=0.402 mm

(a)The average wavelength of the sunlight (in nm) = 550nm

(b) The diameter of the pinhole (in mm). D=0.402mm

What will be the average wavelength of sunlight and the diameter of the pinhole?

We know that the visible range of sunlight is 400 nm to 700 nm.

So the average wavelength

[tex]\lambda _{avg}=\dfrac{\lambda_1+ \lambda_2}{2}[/tex]

[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]

[tex]\lambda_{avg} = 550 nm[/tex]

To find out  the diameter

We know that the relation

[tex]D=\dfrac{2.44\lambda{avg}L}{w}[/tex]

Where D=Diameter of pinhole

λ=Average wavelength

w=Central distance

L=length

So now by putting the values

[tex]D=\dfrac{2.44\times550\times10^{-9}\times 3}{0.01}[/tex]

[tex]D=0.402mm[/tex]

Thus

(a)The average wavelength of the sunlight (in nm) = 550nm

(b) The diameter of the pinhole (in mm). D=0.402mm

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