Question

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:

• A. \(40\ \text{m}\)
• B. \(30\ \text{m}\)
• C. \(20\ \text{m}\)
• D. \(10\ \text{m}\)

Hint:

For ray optics approximation, use the distance estimate \(L\approx \frac{a^2}{\lambda}\).

Answer 1

The Correct Option is B

Concept:
Ray optics is a good approximation when diffraction effects are negligible. A commonly used distance estimate is: \[ L \approx \frac{a^2}{\lambda} \] where, \[ a=\text{aperture size} \] \[ \lambda=\text{wavelength} \]

Step 1: Convert aperture into metre.
\[ a=3\ \text{mm} \] \[ a=3\times 10^{-3}\ \text{m} \]

Step 2: Convert wavelength into metre.
\[ \lambda=300\ \text{nm} \] \[ \lambda=300\times 10^{-9}\ \text{m} \] \[ \lambda=3\times 10^{-7}\ \text{m} \]

Step 3: Apply the formula.
\[ L=\frac{a^2}{\lambda} \] \[ L=\frac{(3\times 10^{-3})^2}{3\times 10^{-7}} \] \[ L=\frac{9\times 10^{-6}}{3\times 10^{-7}} \] \[ L=3\times 10^{1} \] \[ L=30\ \text{m} \]

Step 4: Final conclusion.
Hence, the distance is: \[ \boxed{30\ \text{m}} \]