JEE Advance - Physics (2008 - Paper 1 Offline - No. 10)
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).
If $$d=\lambda$$, the screen will contain only one maximum
If $$\lambda < d < 2\lambda$$, at least one more maximum (besides the central maximum) will be observed on the screen
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
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
Explanation
If $$d=\lambda$$, the maximum path difference (path difference is given by $$d \sin\theta$$) will be less than $$\lambda$$. So there will be only central maximum on the screen, because in the equation $$d\sin\theta=n\lambda,n$$ can take only one value.
If $$\lambda < d < 2\lambda$$, then the maximum path difference will be less than 2$$\lambda$$. So there will be two more maximum on the screen in addition to the central maximum.
Intensity of dark fringes becomes zero when intensities at the two slits are equal. Initial intensity at both the slits is unequal so there will some brightness at dark fringes. Hence, when intensity of both slits is made equal, the intensity at dark fringes on screen will reduces to zero.
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