A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens. A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.

Explanation:

The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.

In compound microscopes, the distance between the two lenses is expressed as L = v0+ue

v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.

Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.

Given L = 18.0cm

Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm

1/f0 = 1/u0+1/v0

f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.

1/v0 = 1/7.4-1/12

1/v0 = 0.1351-0.0833

1/v0 = 0.0518

v0 = 1/0.2184

v0 = 19.31cm

Note that v0 = ue = 19.31cm

To get ve, we will use the lens formula 1/fe = 1/ue+1/ve

1/ve = 1/fe-1/ue

Given ue = 19.31cm and fe = -7.00cm

1/ve = -1/7.0-1/19.31

1/ve = -0.1429-0.0518

1/ve = -0.1947

ve = 1/-0.1947

ve = -5.14cm

Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens

b) Magnification of the final image M = ve/ue

M = 5.14/19.31

M = 0.27

Magnification of the final image is 0.27

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