FIX THIS:The distance between the slits in young's double slit experiment is 0.25cm. Interference fringes are formed on a screen placed at a distance of 100cm from slits. The distance of the third dark fringe from the central bright fringe is 0...59cm. What is the wavelength of the incident light?

Correct answer: B. 590nm

  • A. 390nm
  • B. 590nm
  • C. 690nm
  • D. 790nm
  • E. 990nm

Explanation

Given:Distance between the slits (d) = 0.25 cm = 0.0025 mDistance to the screen (D) = 100 cm = 1 mDistance of the third dark fringe (y3) = 0.59 cm = 0.0059 mFor dark fringes in Young's experiment, the position ym is given by:ym=(m+1/2)λD/dλ=y3*d/(m+1/2)Dλ=0.0000059m=590nm

Written and checked by , contributorLast updated
Report an error

The more specific you are, the faster it gets fixed. A source beats an opinion.

Prefer email? [email protected]

About Optics

Light is studied through reflection, refraction, dispersion, interference, diffraction and polarization, linking ray behavior with its wave nature. Questions may involve mirrors, lenses, optical instruments, refractive index and image formation, while geometrical optics deals with rays and wave optics explains effects such as interference and diffraction.

Practise Optics

665 free Optics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Discussion

Stuck on an option, or know a faster way to get there? Ask or explain it here.

No comments yet. Be the first to explain this one.

Exams that ask Physics questions like this

Physics is on 11 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.

More Optics questions