1. How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled? Refer to the diffraction grating equations in your answer.

a) The distance between the ineas doubles, intensity decreases with distance

b) The distance between the ineas doubles

Explanation:

The diffraction pattern of a grid is given as a percentage

                d sin θ = m λ

where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.

Let's apply this to our case

a) distance from grid to observation screen doubles

Here we have two effect:

* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance

* The other factor can be found using trigonometry

          tan θ= y / L

where y is the distance from the central maximum to the line under study and L is the distance to the screen

In general, diffraction experiments cover very small angles

             tan θ = sin θ/ cos θ = sin θ

we substitute

          sinθ = y / L

we subtitle into the diffraction equation

          d y / L = m λ

          y = L / d m λ

          L = 2 L₀

          y = 2 L₀ m λ / d

we see that by doubling the distance to the screen the lines we are seeing are separated by double

b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half

          d = d₀o / 2

          y = (L m λ) / d

          y = (L m λ/ d₀) 2

we see that The distance between the ineas doubles

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.