Home
Class 12
PHYSICS
In Youngs double slit experiment the 7th...

In Youngs double slit experiment the 7th maximum with wavelength `lambda_1` is at a distance `d_1` and that with wavelength `lambda_2` is at distance `d_2`. Then `d_1//d_2=`

A

`lambda_1//lambda_2`

B

`lambda_2//lambda_1`

C

`lambda_1^2//lambda_2^2`

D

`lambda_2^2//lambda_1^2`.

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

In Young’s Double slit experiment with sodium vapour lamp of wavelength 589 nm and slits 0.589 mm apart, the angular width of central maximum is

In the Youngs double slit experiment, the distance of pth dark fringe from the central maximum

In Youngs double slit experiment, a maximum is obtained when the path difference between the interfering waves is : (Given that m = 1, 2, 3, 4 )

Youngs double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1.10 mm and the screen is placed at a distance of 1 m. The distance between the consecutive bright and dark fringes is

In Young’s double slit experiment, carried out with light of wave length lambda = 5000 overset@A , the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 . The third maximum (taking central maximum as zeroth maximum) will be at what value of x ?