Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.

a)0.51°

b)1.47×10^-4m

Explanation:

a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.

bsin(θ)= mλ(*)

Where b= width of the slit

The distance on the screen from Central angle can be expressed as

Sin(θ)= y/d (**)

d and y is the horizontal distance between slit and screen

If we input eqn(**) into equation (*) we have

y= mλd/b(z)

In order to find angle (θ) we have

(θ)= sin-(1.99×10^-2)/2.25

= 0.51°

Therefore, angle of diffraction θ of the second minimum is 0.51°

b)to find the width of the sloth using eqn(z) by substitute the values, we have

b= (2)(649×10^-9)(2.25)/1.99×10^-2

b= 1.47×10^-4m

Therefore, the width of the slit is 1.47×10^-4m

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

e/c, where e is its total energy