9) An object with a height of 18 cm is placed in front of a converging lens. The image has a height of –9.0 cm.
a) What is the magnification of the lens?
b) If the focal length of the lens is 6.0 cm, how far in front of the lens is the object?
c) Where does the image appear?

Explanation:

a) Magnification = image height / object height = -9 / 18 = -0.5

b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

1/focal length = 1/image distance + 1/object distance

1/6 = 1/(0.5 object distance) + 1/object distance

object distance = 18.0 cm

c) Image appears behind the lens.

Explanation:

mag = ht i/ht o = -9/18 = -1/2

mag = -1/2 = -di/do

do = 2di

put in eqn

1/f = 1/di + 1/do

1/6 = 1/di + 1/2di

2di = 18

do = 18cm

image is behind lens

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

answer is slate

hope this