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In a Young's double slit experiment, a small detector measures an intensity of illumination of I units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be
- a)2I
- b)I
- c)I/4
- d)I/2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a Young's double slit experiment, a small detector measures an ...
If the intensity of central maxima is I , then intensity due to individual source must be I/4.
So if one of the slit is closed then due to remaining slit, intensity will be I/4
Most Upvoted Answer
In a Young's double slit experiment, a small detector measures an ...
Young's Double Slit Experiment: Intensity Measurement
Explanation:
In Young's double slit experiment, a coherent light source is used to produce a pattern of bright and dark fringes on a screen. The pattern is a result of the interference of light waves from two identical slits.
When the two slits are open, the interference pattern is observed. At the centre of the pattern, the intensity of light is maximum, and it is given by the formula:
I = 4I₀ cos²(πdλ/D)
where I₀ is the intensity of light from a single slit, d is the distance between the two slits, λ is the wavelength of light, and D is the distance between the screen and the slits.
If one of the slits is covered, the intensity of light at the centre of the pattern will change. To find the new intensity, let's consider two cases:
Case 1: When one slit is covered, the intensity of light is measured at the centre of the pattern.
In this case, only one slit is contributing to the interference pattern, and the intensity of light at the centre of the pattern is given by the formula:
I' = I₀ cos²(πdλ/D)
Since only one slit is open, the intensity of light is reduced by a factor of 1/2.
Case 2: When one slit is covered, the intensity of light is measured at a point on the screen away from the centre.
In this case, only one slit is contributing to the interference pattern, and the intensity of light at the point is given by the formula:
I' = I₀ cos²(πdλsinθ/D)
where θ is the angle between the line connecting the slit and the point and the line perpendicular to the screen.
Since the intensity of light at the point is a function of the angle θ, it will be different from the intensity at the centre of the pattern.
Therefore, option (c) I/4 is the correct answer as the intensity of light will be reduced by a factor of 1/4 when one of the two (identical) slits is covered.
Explanation:
In Young's double slit experiment, a coherent light source is used to produce a pattern of bright and dark fringes on a screen. The pattern is a result of the interference of light waves from two identical slits.
When the two slits are open, the interference pattern is observed. At the centre of the pattern, the intensity of light is maximum, and it is given by the formula:
I = 4I₀ cos²(πdλ/D)
where I₀ is the intensity of light from a single slit, d is the distance between the two slits, λ is the wavelength of light, and D is the distance between the screen and the slits.
If one of the slits is covered, the intensity of light at the centre of the pattern will change. To find the new intensity, let's consider two cases:
Case 1: When one slit is covered, the intensity of light is measured at the centre of the pattern.
In this case, only one slit is contributing to the interference pattern, and the intensity of light at the centre of the pattern is given by the formula:
I' = I₀ cos²(πdλ/D)
Since only one slit is open, the intensity of light is reduced by a factor of 1/2.
Case 2: When one slit is covered, the intensity of light is measured at a point on the screen away from the centre.
In this case, only one slit is contributing to the interference pattern, and the intensity of light at the point is given by the formula:
I' = I₀ cos²(πdλsinθ/D)
where θ is the angle between the line connecting the slit and the point and the line perpendicular to the screen.
Since the intensity of light at the point is a function of the angle θ, it will be different from the intensity at the centre of the pattern.
Therefore, option (c) I/4 is the correct answer as the intensity of light will be reduced by a factor of 1/4 when one of the two (identical) slits is covered.
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In a Young's double slit experiment, a small detector measures an intensity of illumination of I units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will bea)2Ib)Ic)I/4d)I/2Correct answer is option 'C'. Can you explain this answer?
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