(B) In Fresnel diffraction, the source of light and the screen are at finite distances from the obstacle or aperture. No lenses are required to make t

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

the number of half-period zones that superpose at the point

(B) In Fresnel diffraction, the source of light and the screen are at finite distances from the obstacle or aperture. No lenses are required to make the rays parallel. The resulting diffraction pattern at any point on the screen is determined by the superposition of wavelets originating from different half-period zones of the wavefront. The intensity at a specific point depends on the number of half-period zones that contribute to the superposition at that point, as these zones can interfere constructively or destructively.

Class 12 Physics Wave Optics Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

Medium

Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

TS EAMCET 2013 All TS EAMCET

AP EAMCET 2013 All AP EAMCET

A
the focal length of lens used for observation
B
the number of half-period zones that superpose at the point
C
the square of the sum of the number of half period zones
D
the thickness of the obstacle

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Wave Optics

Subtopic

Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

Previous Year

TS EAMCET 2013 Paper All TS EAMCET years →

Previous Year

AP EAMCET 2013 Paper All AP EAMCET years →

In Fraunhofer diffraction,light of wavelength $6328 \ \mathring{A}$ is incident perpendicularly on a slit of width $0.2 \ mm$. The angular width of the secondary maxima or minima will be: (in $^{\circ}$)

Medium

View Solution

Consider the following statements $A$ and $B$ and identify the correct answer:
$A$. Fresnel's diffraction pattern occurs when the source of light or the screen on which the diffraction pattern is seen, or when both are at a finite distance from the aperture.
$B$. Diffracted light can be used to estimate the helical structure of nucleic acids.

MediumTS EAMCET 2004

View Solution

$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is

MediumAP EAMCET 2020

View Solution

$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

DifficultKCET 2007

View Solution

In a Fraunhofer diffraction of a single slit, when a slit is illuminated by a light of wavelength $6480 \text{ Å}$, the angular width of the central maximum is measured. When the slit is illuminated by light of another wavelength '$\lambda$',the angular width decreases by $25\%$. The value of '$\lambda$' in $\text{Å}$ is:

DifficultMHT CET 2026

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

Similar Questions

In Fraunhofer diffraction,light of wavelength $6328 \ \mathring{A}$ is incident perpendicularly on a slit of width $0.2 \ mm$. The angular width of the secondary maxima or minima will be: (in $^{\circ}$)

$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is

$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module