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If the two slits in Young's experiment have width ratio 1:4, the ratio of intensity at maxima and minima in the interference pattern in
- a)3 : 1
- b)1 : 3
- c)9 : 1
- d)1 : 9
Correct answer is option 'C'. Can you explain this answer?
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The Young's double-slit experiment demonstrates the wave nature of light and the phenomenon of interference. In this experiment, light passes through two narrow slits and creates an interference pattern on a screen placed behind the slits. The intensity of the light at different points on the screen depends on the phase relationship between the waves from the two slits.
Given that the two slits have a width ratio of 1:4, it means that one slit is four times wider than the other. Let's consider the wider slit as S1 and the narrower slit as S2.
1. Intensity at Maxima:
At the maxima of the interference pattern, constructive interference occurs, which means that the waves from the two slits are in phase. The intensity at the maxima is given by the formula:
I_max = 4I1 + 4I2 + 2√(4I1 × 4I2) × cos(Δφ)
where I1 and I2 are the intensities of S1 and S2, respectively, and Δφ is the phase difference between the waves from the two slits.
Since the width ratio of the slits is 1:4, we can assume I1 = I and I2 = 4I, where I is some arbitrary intensity. Substituting these values into the formula, we get:
I_max = 4I + 16I + 2√(4I × 16I) × cos(Δφ)
I_max = 20I + 8√(64I^2) × cos(Δφ)
I_max = 20I + 64I × cos(Δφ)
I_max = 84I + 64I × cos(Δφ)
2. Intensity at Minima:
At the minima of the interference pattern, destructive interference occurs, which means that the waves from the two slits are out of phase. The intensity at the minima is given by the formula:
I_min = 4I1 + 4I2 - 2√(4I1 × 4I2) × cos(Δφ)
Substituting the values of I1 and I2, we get:
I_min = 4I + 16I - 2√(4I × 16I) × cos(Δφ)
I_min = 20I - 8√(64I^2) × cos(Δφ)
I_min = 20I - 64I × cos(Δφ)
I_min = 84I - 64I × cos(Δφ)
3. Ratio of Intensity at Maxima and Minima:
To find the ratio of intensity at maxima and minima, we divide the equation for I_max by the equation for I_min:
(I_max / I_min) = (84I + 64I × cos(Δφ)) / (84I - 64I × cos(Δφ))
Since cos(Δφ) can take values from -1 to 1, we can simplify the ratio as:
(I_max / I_min) = (84I + 64I × cos(Δφ)) / (84I - 64I × cos(Δφ))
(I_max / I_min) = (84 + 64
Given that the two slits have a width ratio of 1:4, it means that one slit is four times wider than the other. Let's consider the wider slit as S1 and the narrower slit as S2.
1. Intensity at Maxima:
At the maxima of the interference pattern, constructive interference occurs, which means that the waves from the two slits are in phase. The intensity at the maxima is given by the formula:
I_max = 4I1 + 4I2 + 2√(4I1 × 4I2) × cos(Δφ)
where I1 and I2 are the intensities of S1 and S2, respectively, and Δφ is the phase difference between the waves from the two slits.
Since the width ratio of the slits is 1:4, we can assume I1 = I and I2 = 4I, where I is some arbitrary intensity. Substituting these values into the formula, we get:
I_max = 4I + 16I + 2√(4I × 16I) × cos(Δφ)
I_max = 20I + 8√(64I^2) × cos(Δφ)
I_max = 20I + 64I × cos(Δφ)
I_max = 84I + 64I × cos(Δφ)
2. Intensity at Minima:
At the minima of the interference pattern, destructive interference occurs, which means that the waves from the two slits are out of phase. The intensity at the minima is given by the formula:
I_min = 4I1 + 4I2 - 2√(4I1 × 4I2) × cos(Δφ)
Substituting the values of I1 and I2, we get:
I_min = 4I + 16I - 2√(4I × 16I) × cos(Δφ)
I_min = 20I - 8√(64I^2) × cos(Δφ)
I_min = 20I - 64I × cos(Δφ)
I_min = 84I - 64I × cos(Δφ)
3. Ratio of Intensity at Maxima and Minima:
To find the ratio of intensity at maxima and minima, we divide the equation for I_max by the equation for I_min:
(I_max / I_min) = (84I + 64I × cos(Δφ)) / (84I - 64I × cos(Δφ))
Since cos(Δφ) can take values from -1 to 1, we can simplify the ratio as:
(I_max / I_min) = (84I + 64I × cos(Δφ)) / (84I - 64I × cos(Δφ))
(I_max / I_min) = (84 + 64
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If the two slits in Young's experiment have width ratio 1:4, the ratio of intensity at maxima and minima in the interference pattern ina)3 : 1b)1 : 3c)9 : 1d)1 : 9Correct answer is option 'C'. Can you explain this answer?
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