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The intensity at the maximum in Young's double slit experiment is I0. Distance between two slits is d = 5λ, where λ is the wavelength of light used is the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?
  • a)
    I0/2
  • b)
    I0
  • c)
    I0/4
  • d)
    3I0/4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The intensity at the maximum in Young's double slit experiment is ...
Given, 
Maximum intensity = Io
Distance between the slits, d = 5λ
Distance of screen from the slit, D = 10d
Using the formula,
Path difference, 
Here,

D = 10d = 50λ
So,

Corresponding phase difference will be



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Most Upvoted Answer
The intensity at the maximum in Young's double slit experiment is ...
Calculation of Intensity in Front of One of the Slits:
Intensity at the maximum in Young's double-slit experiment is given by I0.

Given:
- Distance between two slits (d) = 5
- Wavelength of light used () =
- Distance to screen (D) = 10d

Formula:
- Intensity at a point on the screen in Young's double-slit experiment is given by:
I = I0 * cos²(πdsinθ / )
- For the point in front of one of the slits, θ = 0
- Therefore, sinθ = 0
- Using the above information and formula, we can calculate the intensity in front of one of the slits at a distance D = 10d:
I = I0 * cos²(0)
I = I0 * 1/2
I = I0/2
Therefore, the intensity in front of one of the slits on the screen placed at a distance D = 10d is I0/2, which corresponds to option 'A'.
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The intensity at the maximum in Young's double slit experiment is I0. Distance between two slits isd = 5λ, where λis the wavelength of light used is the experiment. What will be the intensity in front ofone of the slits on the screen placed at a distance D = 10d ?a)I0/2b)I0c)I0/4d)3I0/4Correct answer is option 'A'. Can you explain this answer?
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