In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the…

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),
  1. If $d=\lambda$, the screen will contain only one maximum
  2. If $\lambda < d < 2 \lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
  3. 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
  4. 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

Solution

For $d=\lambda$, there will be only one, acentral maxima. For $\lambda < d < 2 \lambda$, there will be three maximas on the screen corresponding to path difference, $\Delta x=0$ and $\Delta x=\pm \lambda$ correct options are (a) and (b).

Asked in: JEE Advanced 2008 (Paper 1)

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