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  >  A parallel beam of light of wavelength &lambd... Start Learning for Free
SaveJoin the Discussion on the App
A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is       [NEET Kar. 2013]
  • a)
    πλ
  • b)
    2π
  • c)
    3π
  • d)
    4π
Correct answer is option 'D'. Can you explain this answer?
FREE This question is part of
Download PDF Attempt Test
Download PDF Attempt this Test
Verified Answer
A parallel beam of light of wavelength λ is incident normally o...
Conditions for diffraction minima are
Path diff. Δx = nλ and Phase diff. δφ = 2nπ
Path diff. = nλ = 2λ
Phase diff. = 2nπ = 4π (∵n = 2)
View all questions of this test
Most Upvoted Answer
A parallel beam of light of wavelength λ is incident normally o...

Calculation of Phase Difference:
Phase difference between the rays coming from the two edges of the slit can be calculated using the formula:

\[\text{Phase difference} = \frac{2\pi d \sin\theta}{\lambda}\]

Given:
At the second minimum of the diffraction pattern, the phase difference is 2π.

Using the formula:
\[\frac{2\pi d \sin\theta}{\lambda} = 2\pi\]

Solving for λ:
\[d \sin\theta = \lambda\]

Explanation:
At the second minimum, the path difference between the two rays is half of the wavelength. This leads to destructive interference at that point. Therefore, the phase difference between the rays is 4π as the phase change of 2π corresponds to a path difference of one wavelength.

Therefore, the correct answer is option 'D' (4π).
< Previous Question Next Question >
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. The minimum and maximum frequencies in the musical programme were 550 Hz and 10 kHz. Which frequency was better audible around the pillar obstacle?

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

A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer?
Question Description
A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? 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 A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is [NEET Kar. 2013]a)πλb)2πc)3πd)4πCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
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