All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
Class 12 Exam  >  Class 12 Questions  >  If diffraction occurs through a single slit t... Start Learning for Free
If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima?
Clear all your Class 12 doubts with EduRev
Most Upvoted Answer
If diffraction occurs through a single slit then intensity of first se...
Diffraction through a Single Slit

Diffraction is the bending of waves around an obstacle or through an aperture.

When a monochromatic light wave passes through a single slit, it diffracts and produces a diffraction pattern on a screen.

The central maximum is the brightest spot on the screen, and the intensity of the light decreases as we move away from the central maximum.

Intensity of First Secondary Maxima

The first secondary maxima is the first bright spot on either side of the central maximum.

The intensity of the first secondary maxima is not zero, but it is much less than the intensity of the central maximum.

The intensity of the first secondary maxima is approximately 4% of the intensity of the central maximum.

This means that the central maximum is about 25 times brighter than the first secondary maxima.

Explanation

The intensity of the light at a point on the screen is proportional to the square of the amplitude of the wave at that point.

The amplitude of the wave is a function of the angle of diffraction and the slit width.

The angle of diffraction determines the position of the bright spots on the screen, while the slit width determines the width of the bright spots.

The first secondary maxima is produced by waves that interfere constructively after passing through the slit at a certain angle.

The angle of diffraction for the first secondary maxima is given by sinθ = λ/d, where λ is the wavelength of the light and d is the width of the slit.

The intensity of the first secondary maxima can be calculated using the single slit diffraction equation, which is given by I(θ) = (I_0)(sin(α)/α)^2, where I_0 is the intensity of the incident wave and α = (πd/λ)sinθ.

Conclusion

In conclusion, the intensity of the first secondary maxima is much less than the intensity of the central maximum, and it is approximately 4% of the intensity of the central maximum. This means that the central maximum is about 25 times brighter than the first secondary maxima.
Community Answer
If diffraction occurs through a single slit then intensity of first se...
A)4percent
b)25percent
c)75percent
d)50percen



correct ans.:1
Explore Courses for Class 12 exam

Similar Class 12 Doubts

Read the following text and answer the following questions on the basis of the same:Diffraction in a hall:A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half.Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly.B was waiting outside the closed door. The door was not fully closed. There was a little opening.But surprisingly, A could hear the music programme.This happened due to diffraction of sound.The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound.Diffraction in such cases helps the sound to "bend around" the obstacles.In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies.B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.Q. Diffraction of sound takes place more when

Read the following text and answer the following questions on the basis of the same:Diffraction in a hall:A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half.Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly.B was waiting outside the closed door. The door was not fully closed. There was a little opening.But surprisingly, A could hear the music programme.This happened due to diffraction of sound.The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound.Diffraction in such cases helps the sound to "bend around" the obstacles.In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies.B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.Q. Diffraction is more pronounced with ______ wavelengths.

Read the following text and answer the following questions on the basis of the same:Diffraction in a hall:A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half.Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly.B was waiting outside the closed door. The door was not fully closed. There was a little opening.But surprisingly, A could hear the music programme.This happened due to diffraction of sound.The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound.Diffraction in such cases helps the sound to "bend around" the obstacles.In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies.B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.Q. A and B could hear the music programme due to phenomenon named

Read the following text and answer the following questions on the basis of the same:Diffraction in a hall:A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half.Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly.B was waiting outside the closed door. The door was not fully closed. There was a little opening.But surprisingly, A could hear the music programme.This happened due to diffraction of sound.The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound.Diffraction in such cases helps the sound to "bend around" the obstacles.In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies.B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.Q. The minimum and maximum frequencies in the musical programme were 550 Hz and 10 kHz. Which frequency was better audible around the pillar obstacle?

If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima?
Question Description
If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima?.
Solutions for If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? defined & explained in the simplest way possible. Besides giving the explanation of If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima?, a detailed solution for If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? has been provided alongside types of If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? theory, EduRev gives you an ample number of questions to practice If diffraction occurs through a single slit then intensity of first secondary maxima becomes----percentage of central maxima? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
SaveJoin the Discussion on the App for FREE
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup