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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?
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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)
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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).
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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?
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