Question

The distance for which ray optics becomes a good approximation for an aperture of 0.3 cm and a light of wavelength 6000 $\text{\AA}$ is

• A. 12 m
• B. 15 m
• C. 24 m
• D. 30 m

Hint:

The Fresnel distance $z_F = a^2/\lambda$ gives the maximum distance for which the effects of diffraction are negligible, and geometric optics (ray optics) holds true. Beyond this distance, diffraction must be considered.

Answer 1

The Correct Option is B

Step 1: State the condition for the validity of ray optics.
Ray optics (or geometric optics) is a good approximation as long as the size of the aperture is much larger than the wavelength of light, and the distance traveled ($z$) is less than the Fresnel distance. The Fresnel distance ($z_F$) is the maximum distance for which ray optics is a good approximation. Beyond this distance, the wave nature of light (diffraction) becomes significant (Fraunhofer diffraction region). The formula for the Fresnel distance is: \[ z_F = \frac{a^2}{\lambda}. \] where $a$ is the size of the aperture and $\lambda$ is the wavelength of light.

Step 2: Convert the given values to standard units (SI).
Aperture size: $a = 0.3 \text{ cm} = 0.3 \times 10^{-2} \text{ m} = 3 \times 10^{-3} \text{ m}$.
Wavelength: $\lambda = 6000 \text{ \AA}$. Since $1 \text{ \AA} = 10^{-10} \text{ m}$. \[ \lambda = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}. \]

Step 3: Calculate the Fresnel distance $z_F$.
Substitute the values into the formula: \[ z_F = \frac{(3 \times 10^{-3} \text{ m})^2}{6 \times 10^{-7} \text{ m}}. \] \[ z_F = \frac{9 \times 10^{-6} \text{ m}^2}{6 \times 10^{-7} \text{ m}}. \] \[ z_F = \left(\frac{9}{6}\right) \times 10^{(-6 - (-7))} \text{ m} = 1.5 \times 10^1 \text{ m}. \] \[ z_F = 15 \text{ m}. \] The distance for which ray optics is a good approximation is less than or equal to the Fresnel distance. The value itself is 15 m.