Question:

Fresnel distance for an aperture of size a illuminated by a parallel beam of light of wavelength λ, deciding the validity of ray optics is

Updated On: Apr 10, 2025
  • $\frac{λ}{a^2}$
  • $λa$
  • $\frac{a^2}{λ}$

  • $a^2λ$

  • $a^2λ^2$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Approach Solution - 1

The Fresnel distance (\(D_f\)) is the distance at which diffraction effects become significant and the assumptions of ray optics (such as the approximation of light as straight lines) break down. It is determined by the wavelength of light (\( \lambda \)) and the size of the aperture (\( a \)) through which the light passes.

The formula for the Fresnel distance is given by:

\[ D_f = \frac{a^2}{\lambda} \]

Where:

  • \( a \) is the size of the aperture,
  • \( \lambda \) is the wavelength of the light.

This formula shows the relationship between the aperture size and the wavelength in determining the point at which diffraction effects become important.

Correct Answer:

Correct Answer: (C) \( \frac{a^2}{\lambda} \)

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

The Fresnel distance (also known as the Fresnel number criterion) is the distance beyond which ray optics becomes valid. It is derived by comparing the spreading of the wavefront due to diffraction with the size of the aperture.

Let: 

  • \( a \) = size (or radius) of the aperture
  • \( \lambda \) = wavelength of the light

The formula for Fresnel distance \( z_F \) is: \[ z_F = \frac{a^2}{\lambda} \]

Explanation: At distances much larger than \( z_F \), diffraction effects are negligible, and geometrical (ray) optics can be applied reliably. For distances smaller than \( z_F \), wave nature of light and diffraction must be considered.

Correct Answer: \[ \boxed{\frac{a^2}{\lambda}} \]

Was this answer helpful?
0
0

Top Questions on Ray optics and optical instruments

View More Questions