
A monochromatic light of wavelength $700$nm is incident on a $3.5$mm wide aperture. Find the distance up to which the ray of light can travel so that its spread is less than the size of aperture?
Answer
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Hint: We will use Fresnel distance to describe the length to which the spread becomes comparable to the aperture width or not. It is denoted by $Z_{f}$. It gives an excellent approach to ray optics. For distances <<< $Z_{f}$, spread due to diffraction is smaller than the width aperture. So, we have to find the Fresnel distance.
Formula used: $Z_{f} = \dfrac{a^{2}}{\lambda}$
where, $Z_{f}$ is Fresnel distance.\\
a = width of the aperture.\\
λ = wavelength of light.\\
Complete answer:
We will use 1mm = $10^{-3}$m and 1nm = $10^{-9}$m
Given: - size of aperture, $a = 3.5 \ mm= 3.5 \times 10^{-3}$m
Wavelength of monochromatic light, $\lambda$ = $700 \times 10^{-9}$m.
We will use the formula for finding the distance so that the spread of light rays is smaller than the size of aperture.
$Z_{f} = \dfrac{a^{2}}{\lambda}$
where, a = size of aperture or aperture width.
λ = wavelength of monochromatic light.
Now, we put the values of aperture width, a and the wavelength of light, λ.
We will get,
$Z_{f} = \dfrac{a^{2}}{\lambda} = \dfrac{3.5 \times 10^{-3}m }{700 \times 10^{-9}m}= 17.5\ $m
So, $17.5\ $m is the distance up to which the ray of light can travel so that its spread is less than the size of aperture.
Additional Information:
The distance travelled by light waves for the spread to occur equal to the angular spread due to the diffraction or the Distance of the screen from the slit, so that the spreading of light due to diffraction from the center of the screen is just equal to the size of the slit, is called fresnel distance.
Note:
Answer should be in SI unit of length i.e., meter. So, change the units of given value accordingly. For distances<<< $Z_{f}$ , is the distance to which ray optics is a good approximation. For distances >>> $Z_{f}$ Spread due to the diffraction controls over ray optics and wave optics will become valid.
Formula used: $Z_{f} = \dfrac{a^{2}}{\lambda}$
where, $Z_{f}$ is Fresnel distance.\\
a = width of the aperture.\\
λ = wavelength of light.\\
Complete answer:
We will use 1mm = $10^{-3}$m and 1nm = $10^{-9}$m
Given: - size of aperture, $a = 3.5 \ mm= 3.5 \times 10^{-3}$m
Wavelength of monochromatic light, $\lambda$ = $700 \times 10^{-9}$m.
We will use the formula for finding the distance so that the spread of light rays is smaller than the size of aperture.
$Z_{f} = \dfrac{a^{2}}{\lambda}$
where, a = size of aperture or aperture width.
λ = wavelength of monochromatic light.
Now, we put the values of aperture width, a and the wavelength of light, λ.
We will get,
$Z_{f} = \dfrac{a^{2}}{\lambda} = \dfrac{3.5 \times 10^{-3}m }{700 \times 10^{-9}m}= 17.5\ $m
So, $17.5\ $m is the distance up to which the ray of light can travel so that its spread is less than the size of aperture.
Additional Information:
The distance travelled by light waves for the spread to occur equal to the angular spread due to the diffraction or the Distance of the screen from the slit, so that the spreading of light due to diffraction from the center of the screen is just equal to the size of the slit, is called fresnel distance.
Note:
Answer should be in SI unit of length i.e., meter. So, change the units of given value accordingly. For distances<<< $Z_{f}$ , is the distance to which ray optics is a good approximation. For distances >>> $Z_{f}$ Spread due to the diffraction controls over ray optics and wave optics will become valid.
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