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In Young’s double slit experiment, the fringe width is 1×10⁻⁴ m, if the distance between the slit and screen is doubled and the distance between two slits is reduced to half and the wavelength is changed from 6.4×10⁻⁷ m to 4.0×10⁻⁷ m, then the value of new fringe width will be
Options
(a) 2.5×10⁻⁴ m
(b) 2.0×10⁻⁴ m
(c) 1×10⁻⁴ m
(d) 0.5×10⁻⁴ m
Correct Answer:
2.5×10⁻⁴ m
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Calcium plate has a maximum possible radiation wavelength 400 nm to eject electrons.
- The maximum velocity of a particle, executing simple harmonic motion with an amplitude
- A charge of 40 µC is given to a capacitor having capacitance C=10 µF
- The optical path of a monochromatic light is same if it goes through 4.0 cm
- The value of Planck’s constant is
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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