Adiverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. an object is placed 38 cm to the left of the converging lens. where will the final image be located? express your answer using two significant figures. where will the image be the diverging lens is 45 cm from the converging len? express your answer using two significant figures.

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

0.023 m  right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find  image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v     1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

so

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m  right of diverging 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.