Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart. An object 2.50mm tall is placed 20.0cm to the left of the first (diverging) lens. Note that focal length of diverging lens is a negative number while focal length of converging lens would be positive.

(a) Draw a figure showing both lenses and use principal rays to find approximate position of the image formed by the first lens.
(b) Calculate how far from this first lens is the first image formed.
(c) Calculate how far from this first lens is the final image formed.
(d) What is the height of the final image?
(e) Is it erect or inverted?

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

= - 7.5 cm

first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

magnification by second lens = v / u

= 44 / - 16.5

= - 2.67

d )

total magnification

= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

trig 15.3 cos

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.