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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