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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)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?
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)
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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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