Question:medium

The distance estimation for which ray optics is a good approximation for an aperture of \(3\ \text{mm}\) and wavelength \(300\ \text{nm}\) would be:

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For ray optics approximation, use the distance estimate \(L\approx \frac{a^2}{\lambda}\).
Updated On: May 30, 2026
  • \(40\ \text{m}\)
  • \(30\ \text{m}\)
  • \(20\ \text{m}\)
  • \(10\ \text{m}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Light behaves as a ray (moving in straight lines) only as long as diffraction effects are negligible.
As a beam of light passes through an aperture of size \(a\), it spreads out due to diffraction.
The distance at which this spreading becomes equal to the size of the aperture itself is called the Fresnel Distance (\(Z_f\)).
For distances much smaller than \(Z_f\), ray optics is a valid approximation. At distances comparable to or greater than \(Z_f\), wave optics (diffraction) must be considered.
Step 2: Key Formula or Approach:
The Fresnel distance (\(Z_f\)) is given by the formula:
\[ Z_f = \frac{a^2}{\lambda} \]
Where:
\(a\) = size of the aperture
\(\lambda\) = wavelength of light used
Step 3: Detailed Explanation:
Given values:
Aperture size, \(a = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}\)
Wavelength, \(\lambda = 300 \text{ nm} = 300 \times 10^{-9} \text{ m} = 3 \times 10^{-7} \text{ m}\)

Now, substitute these into the Fresnel distance formula:
\[ Z_f = \frac{(3 \times 10^{-3})^2}{3 \times 10^{-7}} \]
Calculate the numerator:
\[ (3 \times 10^{-3})^2 = 9 \times 10^{-6} \text{ m}^2 \]
Now divide:
\[ Z_f = \frac{9 \times 10^{-6}}{3 \times 10^{-7}} \]
\[ Z_f = \left(\frac{9}{3}\right) \times 10^{-6 - (-7)} \]
\[ Z_f = 3 \times 10^1 \]
\[ Z_f = 30 \text{ m} \]
This means ray optics provides a good approximation for distances up to about 30 meters from the aperture.
Step 4: Final Answer:
The required distance estimation is 30 m.
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