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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?
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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
Using the formula,
Path difference,
Here,
D = 10d = 50λ
So,
Corresponding phase difference will be
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'.
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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