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In Young’s double slit experiment, the slit seperation is 1 mm and the screen is 1 m from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd minima from the central maxima is
Options
(a) 0.50 mm
(b) 1.25 mm
(c) 1.50 mm
(d) 1.75 mm
Correct Answer:
1.25 mm
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The numebr 0 (zero) is required for
- The kinetic energy needed to project a body of mass m from the surface of earth
- When we close one slit in the young’s double slit experiment, then
- A block of mass m moving at a velocity v collides with another block of mass 2m
- The acceleration due to gravity becomes (g/2) where g=acceleration due to gravity
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The numebr 0 (zero) is required for
- The kinetic energy needed to project a body of mass m from the surface of earth
- When we close one slit in the young’s double slit experiment, then
- A block of mass m moving at a velocity v collides with another block of mass 2m
- The acceleration due to gravity becomes (g/2) where g=acceleration due to gravity
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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