Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on the screen. Phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is
I
2I
3I
4I

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): In standard YDSE experiment if upper slit is slightly moved downward then central maxima shifts downward. Reason (R): Fringe width in such case will increase.  
2. In the Young's double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that :- (A) The intensities of individual sources are 5 and 4 units respectively (B) The intensities of individual sources are 4 and 1 units respectively (C) The ratio of their amplitudes is 3 (D) The ratio of their amplitudes is 2
3. Assertion (A): When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of shadow of the obstacle. Reason (R): Constructive interference occurs at the centre of the shadow.
4. Assertion (A): In Young's experiment, the fringe width for dark fringes is different from that for bright fringes. Reason (R): In Young's double slit experiment with a source of white light, only black and white fringes are observed.  
5. Assertion (A): The central fringe is bright or dark, it dependents on the initial phase difference between the two coherent sources. Reason (R): The pattern and position of fringes always remains same even after the introduction of transparent medium in a path of one of the slit.