Light of wavelength 575 nm falls on two double slits spaced 0.30 mm apart. What is the required distance from the slits to a screen if the spacing between the first and second dark fringe is to be 4.00 mm

L = 2.1 m

Explanation:

From Young's Double Slit Experiment, the formula for the distance between two consecutive dark or bright fringes, called fringe spacing, is derived as:

Δx = λL/d

where,

Δx = distance between first and second dark fringe = 4 mm = 4 x 10⁻³ m

λ = wavelength of light = 575 nm = 5.75 x 10⁻⁷ m

d = distance between the slits = 0.3 mm = 3 x 10⁻⁴ m

L = Distance from slits to screen = ?

Therefore,

4 x 10⁻³ m = (5.75 x 10⁻⁷ m)(L)/(3 x 10⁻⁴ m)

L = (4 x 10⁻³ m)/(1.92 x 10⁻³)

L = 2.1 m

friction

explanation:

they us to move forward and backwards

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m