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In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is l. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).
  • a)
    If d = λ, the screen will contain only one maximum
  • b)
    If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen
  • c)
    If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase
  • d)
    If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase
Correct answer is option 'A,B'. Can you explain this answer?
Verified Answer
In a Young’s double slit experiment, the separation between the ...
The condition to obtain maxima in the phenomenon observed in young's double slit experiment is d sin θ = nλ where n is an integer
l sin q = nλ
⇒ sin q = n
When n = 0, θ = 0
When n = 1, θ = 90° (This will be a point on the screen which will be at infinity and therefore not practical)
Other values of n are invalid  as –1 < sin θ <1 .
⇒ The screen will have only one maxima.
The possible values of n are 0,  + 1, –1.
⇒ There is at least one more maxima (besides the central maxima, option [B] is correct.
We know that

Initially I1 = 4 I and I2 = I
∴    Imax = 9 I and Imin = I
When I1 = I2 = I then  Imax  = 4 I and Imin = 0
i.e., when the intensities become equal, Imin reduces to zero.
Options [C] and [D] are incorrect.
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Most Upvoted Answer
In a Young’s double slit experiment, the separation between the ...
Explanation:

Option A:
When the separation between the two slits is equal to the wavelength of light (d = λ), the two waves from the slits will be in phase at all points on the screen, resulting in constructive interference and a single maximum. This is because the path difference between the two waves is zero.

Option B:
If the separation between the slits is between λ and 2λ (λ < d="" />< 2λ),="" there="" will="" be="" additional="" maxima="" and="" minima="" observed="" on="" the="" screen="" due="" to="" interference="" patterns.="" the="" positions="" of="" these="" maxima="" and="" minima="" can="" be="" calculated="" using="" the="" formula="" for="" double-slit="" />

Option C:
If the intensity of light falling on slit 1 is reduced to match that of slit 2, the intensity of the observed fringes will increase. This is because the overall intensity of the interference pattern will be determined by the square of the sum of the amplitudes of the two waves from the slits.

Option D:
If the intensity of light falling on slit 2 is increased to match that of slit 1, the intensities of the observed dark and bright fringes will increase. This is because the overall intensity of the interference pattern will be determined by the square of the sum of the amplitudes of the two waves from the slits.
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In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is l. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).a)If d = λ, the screen will contain only one maximumb)If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screenc)If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increased)If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increaseCorrect answer is option 'A,B'. Can you explain this answer?
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