The question asks for the distance beyond which ray optics is a valid approximation for light passing through an aperture.
Determining Ray Optics Validity
Ray optics is a limit of wave optics where diffraction effects are negligible. This occurs when the wavelength ($\lambda$) is much smaller than the dimensions of the aperture ($a$) and the distance ($d$) to the observation point. A common criterion involves the Fresnel number ($N_F$).
Fresnel Number Criterion
The Fresnel number is defined as $N_F = \frac{a^2}{\lambda d}$.
- When $N_F \gg 1$, Fresnel diffraction is significant, and ray optics is not valid.
- When $N_F \ll 1$, diffraction effects are small, and ray optics provides a good approximation.
We need to find the distance $d_0$ such that for distances $d > d_0$, $N_F < 1$. We can find this threshold distance by setting $N_F = 1$.
Calculating Threshold Distance
Setting the Fresnel number to 1:
$ N_F = \frac{a^2}{\lambda d_0} = 1 $Solving for the threshold distance $d_0$:
$ d_0 = \frac{a^2}{\lambda} $Substituting Values and Finding the Distance
Given:
- Aperture width, $a = 6 \text{ mm} = 6 \times 10^{-3} \text{ m} $
- Wavelength, $\lambda = 6000 \text{ \AA} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} $
Substitute these values into the formula for $d_0$:
$ d_0 = \frac{(6 \times 10^{-3} \text{ m})^2}{6 \times 10^{-7} \text{ m}} $ $ d_0 = \frac{36 \times 10^{-6} \text{ m}^2}{6 \times 10^{-7} \text{ m}} $ $ d_0 = 6 \times 10^1 \text{ m} $ $ d_0 = 60 \text{ m} $Conclusion
Ray optics is sufficiently valid beyond the distance $d_0 = 60 \text{ m}$.
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