For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 18 m
A beam through an aperture of size \( a \) spreads by diffraction through an angle of about \( \lambda/a \). Up to the Fresnel distance this spreading is small compared with the aperture itself, so ray optics is a good approximation.
\( \displaystyle z_F = \frac{a^{2}}{\lambda} \)
\( a = 3\,\mathrm{mm} = 3\times10^{-3}\,\mathrm{m} \)
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( \displaystyle z_F = \frac{(3\times10^{-3})^{2}}{5\times10^{-7}} = \frac{9\times10^{-6}}{5\times10^{-7}} = 18\,\mathrm{m} \)
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 – “… We define a quantity zF called the Fresnel distance by the following equation
\( \displaystyle z_F \simeq a^2/\lambda \qquad (10.31) \)
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.”
The Validity of Ray Optics – “… Example 10.7 For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?
Solution
\( \displaystyle z_F = \frac{a^2}{\lambda} = \frac{\left(3 \times 10^{-3}\right)^2}{5 \times 10^{-7}} = 18\ \mathrm{m} \)”