1.

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).

Answer»

If `d = lambda`, the screen will contain only one maximum.
If `lambda lt d lt 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.

Solution :If `d = lambda`, then 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 lt d lt 2 lambda`, then the maximum path difference will be less than `2 lambda`. So there will be two more maximum on screen in ADDITION to the central maximum. Intensity of the dark fringes becomes zero whenintensititesat the two SLITS are equal.Initial intensity atboth the slits are unequal so there will somebrightnessat dark finge.Hence when intensity of both slits is made same the intensity at dark fringe on screen will decrease to zero. The alternative (c) and (d) are not correct.


Discussion

No Comment Found

Related InterviewSolutions