The distance estimation for which ray optics is good approximation for an aperture of 3 mm and wavelength 300 nm would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30 m
A beam from an aperture of size \( a \) spreads by diffraction through an angle of about \( \lambda/a \). The spread becomes as large as the aperture itself at the Fresnel distance; up to this distance the beam travels almost without spreading, so ray optics is a good approximation.
\( \displaystyle z_F = \frac{a^{2}}{\lambda} \)
\( a = 3\,\mathrm{mm} = 3\times10^{-3}\,\mathrm{m} \)
\( \lambda = 300\,\mathrm{nm} = 3\times10^{-7}\,\mathrm{m} \)
\( \displaystyle z_F = \frac{(3\times10^{-3})^{2}}{3\times10^{-7}} = \frac{9\times10^{-6}}{3\times10^{-7}} = 30\,\mathrm{m} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
The Validity of Ray Optics – “An aperture (i.e., slit or hole) of size a illuminated by a parallel beam sends diffracted light into an angle of approximately ≈ λ/a. … We define a quantity zF called the Fresnel distance by the following equation
\( \displaystyle z_F \simeq a^2/\lambda \)
Equation (10.31) shows that for distances much smaller than zF, the spreading due to diffraction is smaller compared to the size of the beam. … Equation (10.31) also shows that ray optics is valid in the limit of wavelength tending to zero.”