Diffraction and Wave Nature – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): Diffraction is a sure indication of wave nature. Reason (R): Only transverse waves can be diffracted.  
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

Solution Explained:

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:

Assertion (A) is true; diffraction is a characteristic property of all waves. Reason (R) is false; both transverse (e.g., light) and longitudinal (e.g., sound) waves can be diffracted.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): If the phase difference between the light waves emerging from the slits of the Young's experiment is \(\pi\) radian, then central fringe will be dark. Reason (R): Phase difference is equal to \(\frac{2\pi}{\lambda}\) times the effective path difference.  
2. Assertion (A): Two sources of light emit light waves of same frequency but of different amplitudes. Also the phase difference between light waves from the two sources at any point is time independent. Therefore, observable interference will be obtained when light waves from the two sources superimpose. Reason (R): The sources are not coherent due to unequal amplitudes.  
3. 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.  
4. Assertion (A): If two sodium lamps are used illuminating two pinholes, interference fringes will not be observed. Reason (R): Light waves coming from an ordinary source like sodium lamp are unpolarised in nature.  
5. In a Young's double slit experiment the intensity at a point where the path difference is λ/6 (λ being the wavelength of the light used) is I. If I0 denotes the maximum intensity, I/I0 is equal to :