Home
Class 12
PHYSICS
A source emitting light of wavelengths 4...

A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

Text Solution

Verified by Experts

The correct Answer is:
`0.9 mm`

As `lambda lt lt d`
`rArr beta_(1)=(lambda_(1)D)/(d) beta_(2)=(lambda_(2)D)/(d)`
So, `beta_(2)-beta_(1)=((lambda_(2)-lambda_(1))D)/(d) = ((700-580)xx10^(-9)xx1.5)/(0.20xx10^(-3)) =0.9 mm.`
Promotional Banner

Topper's Solved these Questions

  • WAVE OPTICS

    RESONANCE ENGLISH|Exercise Exercise-1 (Part-2)|13 Videos
  • WAVE OPTICS

    RESONANCE ENGLISH|Exercise Exercise-2 (Part-1)|10 Videos
  • WAVE OPTICS

    RESONANCE ENGLISH|Exercise Advanced Level Problems|8 Videos
  • WAVE ON STRING

    RESONANCE ENGLISH|Exercise Exercise- 3 PART II|7 Videos

Similar Questions

Explore conceptually related problems

In a Young's double slit experiment the separation between the slits is 0.10 mm, the wavelength of light used is 600 mm and the interference pattern is observed on a screen 1.0 m away. Find the separation between the successive bright fringes.

Young's interference experiment is performed with monochromatic light. The separation between the slits is 0.500 mm, and the interference pattern on a screen 3.30 m away shows the first side maximum 3.40 mm form the center of the pattern. What is the wavelength ?

In a YDSE green light of wavelength 500 nm is used. Where will be the second bright fringe be formed for a set up in which separation between slits is 4 mm and the screen is placed 1 m from the slits?

Answer the following: (a) In Young's double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen. (b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

Two slits in YDSE are placed 1 mm from each other. Interference pattern is observed on a screen placed 1m from the plane of slits. What is the angular fringe width for a light of wavelength 400 nm

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distance d(gtgtb) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelength are

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and theh screen is at a distance d (gt gtb) from the slits At a point on the screen directly in front of one of the slits, certain wavelengths are missing some of these missing wavelengths are

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distance d (gt gtb) from the slits At a point on the screen directly in front of one of the slits, certain wavelengths are missing some of these missing wavelengths are