Interference Pattern Intensity – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): Interference pattern is obtained on a screen due to two identical coherent sources of monochromatic light. The intensity at the central part of the screen becomes one-fourth if one of the sources is blocked. Reason (R): The resultant intensity at any point is the algebraic sum of the intensities due to two sources.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Young's Double Slit Experiment. Understanding the underlying formula is key to arriving at the correct answer below:

For two identical coherent sources of intensity \(I_0\) each, the central maximum intensity is \(4I_0\). If one source is blocked, the intensity becomes \(I_0\), which is one-fourth of \(4I_0\). The resultant intensity in interference is not an algebraic sum of individual intensities but depends on the phase difference. Thus, A is true and R is false.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. 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.  
2. Assertion (A): The stars which are not resolved in the image produced by the objective of a telescope can't be further resolved by its eye piece. Reason (R): The primary purpose of eyepiece of telescope is to provide the magnification of image produced by the objective.  
3. Assertion (A): In YDSE central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen. Reason (R): In an interference pattern, whatever energy disappears at the minimum, appears at the maximum.  
4. Ratio of intensities of two waves are given by 4 : 1. Then ratio of the amplitudes of the two waves is: (1991)
5. For a parallel beam of monochromatic light of wavelength '$\lambda$', diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be: (2015)