Question:medium

Beyond what distance, the ray optics is sufficiently valid when the aperture is 6 mm wide and the wavelength is 6000 \AA?

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Always convert all given units to SI units (meters) before performing calculations. Remember that \( 1 \text{ \AA} = 10^{-10} \text{ m} \) and \( 1 \text{ mm} = 10^{-3} \text{ m} \).
Updated On: May 28, 2026
  • 50 m
  • 60 m
  • 40 m
  • 10 m
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Ray optics is a simplified model of light propagation that assumes light travels in straight lines.
However, light is a wave and undergoes diffraction when it passes through an aperture.
Diffraction causes the light beam to spread out as it propagates.
The validity of ray optics depends on the distance the light has traveled relative to the size of the aperture and the wavelength of the light.
As long as the diffraction spread is significantly smaller than the aperture width, ray optics is considered a good approximation.
The characteristic distance that defines this boundary is called the Fresnel distance or Fresnel length.
Step 2: Key Formula or Approach:
The Fresnel distance \(Z_F\) is given by the formula:
\[ Z_F = \frac{a^2}{\lambda} \]
where:
\(a\) is the width of the aperture.
\(\lambda\) is the wavelength of the light used.
Step 3: Detailed Explanation:
We are given the following parameters:
Aperture width, \(a = 6 \text{ mm} = 6 \times 10^{-3} \text{ m}\).
Wavelength, \(\lambda = 6000 \text{ \AA} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}\).
To find the distance beyond which ray optics is valid (meaning the distance up to which diffraction effects are negligible), we calculate the Fresnel distance.
Substituting the values into the formula:
\[ Z_F = \frac{(6 \times 10^{-3} \text{ m})^2}{6 \times 10^{-7} \text{ m}} \]
\[ Z_F = \frac{36 \times 10^{-6} \text{ m}^2}{6 \times 10^{-7} \text{ m}} \]
Calculating the numerical coefficients:
\[ \frac{36}{6} = 6 \]
Calculating the powers of ten:
\[ 10^{-6} / 10^{-7} = 10^{-6 - (-7)} = 10^1 \]
Combining these results:
\[ Z_F = 6 \times 10^1 \text{ m} = 60 \text{ m} \]
This means that for distances much less than 60 m, the beam remains roughly the same size as the aperture (ray optics).
At 60 m and beyond, the spread due to diffraction becomes equal to or greater than the aperture size (wave optics).
Step 4: Final Answer:
Ray optics is considered sufficiently valid for distances up to the Fresnel distance, which is calculated as 60 m.
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