Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How far from the center of the pattern are the centers of the first and second dark fringes?

i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards