Covering a Slit
Source: Shmoop Editorial Team. Young's Double Slit Experiment
In a standard double slit experiment, coherent light is passed through 2 slits and diffract. The waves from both slits interfere and produce an interference pattern on the screen that is placed in front of the slits. Assume that the set-up produces an observable interference pattern.
The intensity of the central maximum (Located at the center of the interference pattern) is measured as . A student predicts that if he covers one of the slits, since there is now only one slit that can shine light onto the screen, the intensity at the same location the central fringe was at would be halved ( ).
Is their reasoning correct? If the actual intensity measured upon covering one of the slits is , input as your answer. Assume that single slit interference is negligible.
The answer is 0.25.
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1 solution
First, we need to understand that in classical wave theory, the intensity of a light wave, I is proportional to the square of the amplitude of a wave, A 2 . In mathematical terms, I ∝ A 2 . Thus, using the Principle of Superposition, the resultant amplitude of 2 waves undergoing constructive interference to produce the bright central fringe would be 2 A by denoting the amplitude of a single light wave to be A . By covering one slit, the resultant amplitude of the wave would simply be the amplitude of a single light wave which is A , since only one light wave passes through. Therefore, using the relationship I ∝ A 2 , the intensity must be reduced by a factor of ( 2 1 ) 2 and hence k = 0 . 2 5 .