JEE Challenger
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PhyMedAdv 2008 • Paper 1

Young Double Slit Experiment with Unequal Intensities and Slit Separation Conditions

In a Young's double slit experiment, the separation between the two slits is dd and the wavelength of the light is λ\lambda. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

Options

A

If d=λd = \lambda, the screen will contain only one maximum

Correct
B

If λ<d<2λ\lambda < d < 2\lambda, at least one more maximum (besides the central maximum) will be observed on the screen

Correct
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

Topics & Concepts

Wave OpticsInterferenceYoung's Double Slit

Step-by-Step Solution

To find the correct statements, let us analyze each option individually:

1. Maxima Condition on the Screen (Options A and B):

In Young's double slit experiment, the path difference between light waves reaching an angle θ\theta on a flat screen placed parallel to the plane of the slits is given by:
Δx=dsin⁡θ\Delta x = d \sin\theta
For a flat screen of infinite extent, the angular position θ\theta lies strictly in the range:
−π2<θ<π2  ⟹  −1<sin⁡θ<1-\frac{\pi}{2} < \theta < \frac{\pi}{2} \implies -1 < \sin\theta < 1
Hence, the path difference at any observable point on the screen satisfies:
−d<Δx<d-d < \Delta x < d

The condition for constructive interference (intensity maxima) is:
Δx=nλ,where n∈Z\Delta x = n\lambda, \quad \text{where } n \in \mathbb{Z}
This requires:
−d<nλ<d  ⟹  −dλ<n<dλ-d < n\lambda < d \implies -\frac{d}{\lambda} < n < \frac{d}{\lambda}

  • For Option A (d=λd = \lambda):
    −1<n<1-1 < n < 1
    The only integer value of nn satisfying this inequality is n=0n = 0.
    The values n=±1n = \pm 1 correspond to sin⁡θ=±1\sin\theta = \pm 1 (θ=±90∘\theta = \pm 90^\circ), which would be at infinity and cannot be observed on a flat screen.
    Therefore, only the central maximum (n=0n = 0) is observed on the screen.
    Hence, Option A is correct.

  • For Option B (λ<d<2λ\lambda < d < 2\lambda):
    Since 1<dλ<21 < \frac{d}{\lambda} < 2, the inequality −dλ<n<dλ-\frac{d}{\lambda} < n < \frac{d}{\lambda} is satisfied by:
    n=−1, 0, +1n = -1, \, 0, \, +1
    For n=±1n = \pm 1, the angular positions are given by sin⁡θ=±λd\sin\theta = \pm \frac{\lambda}{d}. Since λd<1\frac{\lambda}{d} < 1, these maxima lie at finite positions on the screen.
    Thus, two additional maxima are observed on the screen, meaning at least one more maximum (besides the central one) is observed.
    Hence, Option B is correct.

2. Intensity of Fringes (Options C and D):

Let the initial intensities be:
I1=4I0andI2=I0I_1 = 4I_0 \quad \text{and} \quad I_2 = I_0
The initial maximum and minimum intensities on the screen are:
Imax=(I1+I2)2=(4I0+I0)2=(2I0+I0)2=9I0I_{\text{max}} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{4I_0} + \sqrt{I_0}\right)^2 = (2\sqrt{I_0} + \sqrt{I_0})^2 = 9I_0
Imin=(I1−I2)2=(4I0−I0)2=(2I0−I0)2=I0I_{\text{min}} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{4I_0} - \sqrt{I_0}\right)^2 = (2\sqrt{I_0} - \sqrt{I_0})^2 = I_0

  • For Option C:
    If I1I_1 is reduced to I0I_0 such that I1′=I2′=I0I_1' = I_2' = I_0:
    Imax′=(I0+I0)2=4I0<9I0(decreases)I_{\text{max}}' = (\sqrt{I_0} + \sqrt{I_0})^2 = 4I_0 < 9I_0 \quad (\text{decreases})
    Imin′=(I0−I0)2=0<I0(decreases)I_{\text{min}}' = (\sqrt{I_0} - \sqrt{I_0})^2 = 0 < I_0 \quad (\text{decreases})
    Therefore, the intensities do not both increase.
    Hence, Option C is incorrect.

  • For Option D:
    If I2I_2 is increased to 4I04I_0 such that I1′=I2′=4I0I_1' = I_2' = 4I_0:
    Imax′=(4I0+4I0)2=16I0>9I0(increases)I_{\text{max}}' = (\sqrt{4I_0} + \sqrt{4I_0})^2 = 16I_0 > 9I_0 \quad (\text{increases})
    Imin′=(4I0−4I0)2=0<I0(decreases)I_{\text{min}}' = (\sqrt{4I_0} - \sqrt{4I_0})^2 = 0 < I_0 \quad (\text{decreases})
    The intensity of the dark fringe decreases to zero rather than increasing.
    Hence, Option D is incorrect.

Conclusion:

The correct options are A and B.

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