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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is (AIEEE 2004)
- a)Infinite
- b)Five
- c)Three
- d)Zero
Correct answer is option 'B'. Can you explain this answer?
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The maximum number of possible interference maxima for slit-separation...
For maxima, d sinθ =nλ
⇒2λsinθ=nλ
2sinθ=n
∴sinθ=n/2
∵sinθ≤1
⇒n/2≤1
n≤2
nmax=2
n=0,1, 2,
So maximum number of possible maxima=5
n=0,±1,±2
⇒2λsinθ=nλ
2sinθ=n
∴sinθ=n/2
∵sinθ≤1
⇒n/2≤1
n≤2
nmax=2
n=0,1, 2,
So maximum number of possible maxima=5
n=0,±1,±2
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The maximum number of possible interference maxima for slit-separation...
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The maximum number of possible interference maxima for slit-separation...
Explanation:
In Young's double-slit experiment, light from a single source is passed through two very narrow slits and is then projected onto a screen. The interference pattern produced on the screen is due to the superposition of the waves from the two slits.
The interference pattern consists of a series of bright fringes (maxima) and dark fringes (minima). The number of interference maxima depends on the distance between the two slits and the wavelength of the light.
When the slit separation is equal to twice the wavelength, the interference pattern becomes more complex, and the number of interference maxima increases. The maximum number of possible interference maxima is given by the formula:
Number of Maxima = (slit separation)/(wavelength) + 1
Substituting the values, we get:
Number of Maxima = (2λ)/(λ) + 1
Number of Maxima = 3
Therefore, the maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is three.
However, in practice, it is difficult to observe more than five interference maxima due to the limitations of the experimental setup.
In Young's double-slit experiment, light from a single source is passed through two very narrow slits and is then projected onto a screen. The interference pattern produced on the screen is due to the superposition of the waves from the two slits.
The interference pattern consists of a series of bright fringes (maxima) and dark fringes (minima). The number of interference maxima depends on the distance between the two slits and the wavelength of the light.
When the slit separation is equal to twice the wavelength, the interference pattern becomes more complex, and the number of interference maxima increases. The maximum number of possible interference maxima is given by the formula:
Number of Maxima = (slit separation)/(wavelength) + 1
Substituting the values, we get:
Number of Maxima = (2λ)/(λ) + 1
Number of Maxima = 3
Therefore, the maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is three.
However, in practice, it is difficult to observe more than five interference maxima due to the limitations of the experimental setup.
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