Home
Class 12
PHYSICS
In YDSE, the two slits are separated by ...

In YDSE, the two slits are separated by `0.1 mm` and they are `0.5 m` from the screen. The wavelenght of light used is `5000A^(0)`. Find the distance between 7th maxima 11th minima on the screen.

Text Solution

AI Generated Solution

To solve the problem of finding the distance between the 7th maxima and the 11th minima on the screen in Young's Double Slit Experiment (YDSE), we can follow these steps: ### Step 1: Convert the given values into standard units - The distance between the slits (d) is given as \(0.1 \, \text{mm}\). We convert this to meters: \[ d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \] ...
Promotional Banner

Topper's Solved these Questions

  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA|54 Videos
  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA (DIFFRACTION)|33 Videos
  • WAVES

    AAKASH SERIES|Exercise EXERCISE-III (Doppler effect :)|15 Videos

Similar Questions

Explore conceptually related problems

In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is 5000 Å. Find the distance between 7th maxima and 11 th minima on the upper side of screen.

In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is 5000 Å. Find the distance between 7th maxima and 11 th mimima on the upper side of screen.

In the youngs double slit experiment ,the slit sepration is 1mm ,the distance between source and screen is 1m ,wavelength of light used is 6500 angstrom.Find the distance between fifth maxima and third minima in (mm)?

In Young's double-slit experiment, the slit separation is 0.5 mm and the screen is 0.5 m away from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

In a YDSE, distance between the slits and the screen is 1m, separation between the slits is 1mm and the wavelength of the light used is 5000nm . The distance of 100^(th) maxima from the central maxima is:

The distace between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm . The screen is 0.5 m away from the slit. If wavelength of the light used is 5000 A^(0) , then the slit width will be

In YDSE the distance between the slits is 1mm and screen is 25nm away from intensities IF the wavelength of light is 6000A the fringe width on the screen is

In YDSE, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the first dark fringe and fourth bright fringe is obtained to be 0.6 cm Determine the wavelength of the light used in the experiment.

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