A) \[{{I}_{0}}\]
B) \[\frac{{{I}_{0}}}{2}\]
C) \[2{{I}_{0}}\]
D) \[4\,{{I}_{0}}\]
Correct Answer: D
Solution :
If you divide the original slit into N strips and represents the light from each strip, when it reaches the screen, by a phasor, then at the central maximum in the diffraction pattern you add N phasors, all in the same direction and each with the same amplitude. The intensity is therefore N2. If you double the slit width, you need 2N phasors, if they are each to have the amplitude of the each to have the amplitude of the phasors you used for the narrow slit. The intensity at the central maximum is proportional to (2N)2 and is, therefore, four times the intensity for the narrow slit.
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