Young’s Double Slit Fringe Width – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): In case of Young double slit experiment width of all fringes were equal. Reason (R): Angular width of fringes were equal.  
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
(4) 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:

In Young's Double Slit Experiment (YDSE), the fringe width \(beta = \frac{\lambda D}{d}\) is constant for all fringes, hence they are of equal width. The angular fringe width \(\theta = \frac{\lambda}{d}\) is also constant. Reason (R) correctly explains Assertion (A).

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): As angle subtended by the diameter of objective lens at the focus of microscope increased, resolving limit also increases. Reason (R): Resolving limit proportional to tangent of the angle subtended by the diameter of objective lens at the focus of microscope.  
2. Assertion (A): If a glass slab is placed in front of one of the slits, then fringe width will decrease. Reason (R): Glass slab will produce no path difference.  
3. Assertion (A): If width of one of the slit in YDSE is slightly increased, then maximum and minimum both Intensity will increase. Reason (R): Intensity reaching from that slit on screen will slightly increase.
4. Assertion (A): In the double slit experiment, if one of the slit is closed, no fringe pattern will be observed on the screen. Reason (R): In interference, phenomenon of diffraction is also included.
5. The Young's double slit experiment is performed with blue and with green light of wavelengths $4360 \AA$ and $5460 \AA$ respectively. If x is the distance of 4th maxima from the central one, then: (1990)