In a two-slit experiment, the slit separation is 3.00 × 10-5 m. the interference pattern is created on a screen that is 2.00 m away from the slits. if the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light?

214.3 x 10⁻⁹ m or 214.3 nm

Explanation:

= slit separation in the experiment = 3.00 x 10⁻⁵ m

= screen distance from the slits = 2.00 m

= Order of the bright fringe = 7

λ = Wavelength of the light

= distance of the bright fringe from the central bright fringe = 10 cm = 0.10 m

Position of the bright fringe is given as

λ = 214.3 x 10⁻⁹ m

λ = 214.3 nm

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.