Physics · Wave Optics · NEET
The Fresnel distance is z_F = a^2 / lambda. Here 'a' is the width of the aperture (or slit) that the beam passes through, and lambda is the wavelength of the light. z_F is a distance in metres. It marks how far the beam can travel and still be treated as a straight ray. Example: a = 3 mm = 3x10^-3 m and lambda = 5x10^-7 m give z_F = (3x10^-3)^2 / (5x10^-7) = 9x10^-6 / 5x10^-7 = 18 m.
A beam passing through an aperture of width 'a' spreads because of diffraction. The half-angular spread is about lambda/a. So the extra spreading in width after travelling a distance z is roughly z x (lambda/a). Ray optics assumes light goes straight, so it only holds while this spreading is smaller than the beam width 'a'. Setting z x (lambda/a) = a gives z = a^2/lambda. Below this distance the beam stays sharp (ray optics works); beyond it the spreading is larger than the beam itself and the wave nature dominates.
At z = z_F, the sideways spread of the beam due to diffraction becomes equal to the original width 'a' of the aperture. Before this point the beam looks like a straight column of light (a shadow with sharp edges). After this point the beam has widened so much that you can no longer treat it as a straight ray. So z_F is the boundary between the ray-optics region and the wave-optics (diffraction) region.
Yes. Because z_F depends on a^2, a wider aperture greatly increases the Fresnel distance. If you double 'a', z_F becomes 4 times larger, so the beam stays straight much farther. This is why a wide beam (like a searchlight) stays a straight column for a long distance, while a very narrow slit spreads out quickly. It also explains why diffraction is hard to notice in daily life: everyday openings (doors, windows) are huge compared to lambda, so z_F is enormous.
They are unrelated quantities. Focal length (f) is a property of a lens or mirror and tells where parallel rays converge. Fresnel distance (z_F) is a property of a beam and aperture, telling how far light travels before diffraction spreading matters. In a PYQ a slit may sit in front of a lens, but z_F = a^2/lambda uses the slit width and wavelength, not the focal length.
Larger. Since z_F = a^2/lambda, a smaller lambda (like blue light or X-rays) gives a bigger Fresnel distance, so the beam stays straight longer and diffraction is weaker. A longer lambda (like red light or radio waves) gives a smaller z_F and spreads sooner. This is why very short wavelengths behave more like straight rays.
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 x 10^-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a distance, so its SI unit is the metre (m). Since z_F = a^2/lambda has units of (m^2)/(m) = m, the answer always comes out in metres when a and lambda are in metres.
Before z_F the beam travels almost straight, so ray optics and the idea of a sharp geometrical shadow work well. After z_F the beam spreads by more than its own width, so diffraction (wave optics) must be used.
Everyday openings are millions of times wider than the wavelength of light. Because z_F = a^2/lambda depends on a^2, the Fresnel distance becomes extremely large, so light behaves like straight rays over normal room distances.
Yes. The half-angular spread due to diffraction is about lambda/a. Multiplying this angle by z_F gives z_F x (lambda/a) = a, which shows that at the Fresnel distance the sideways spread equals the aperture width.
A wider aperture (larger a) and a shorter wavelength (smaller lambda) both increase z_F. Aperture matters most because it is squared in the formula.
