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

(A) If $d=\lambda$, the screen will contain only one maximum; (B) If $\lambda<d<2 \lambda$, 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

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Wave Optics

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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

λ

. 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✓ Correct
(B)If $λ < d < 2 λ$ , 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

Answer: (A), (B)

Solution

For maxima to be observed,

sin θ = \frac{nλ}{d}

must have

∣ sin θ ∣ \leq 1

. If

d = λ

, then

sin θ = n

, which gives only

n = 0

(central maximum) and

n = \pm 1

θ = 9 0^{\circ}

(at infinity), so effectively only one maximum is visible (A is correct). If

λ < d < 2 λ

, then

n = 1

gives

sin θ < 1

, so at least one more maximum exists (B is correct). For interference, maximum intensity is

(I_{1} + I_{2})^{2}

and minimum is

(I_{1} - I_{2})^{2}

. When intensities are equal, contrast improves but the bright fringe intensity may not increase (C and D need careful analysis).

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Wave Optics

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