An object is placed 20 cm away from a curved mirror that has a focal length of 7 cm.
Where is the image located? Is the image magnified or shrunken? Show your work

An image is produced 51.5 cm away from a converging lens. If the object is placed at 32 cm away, where is the focal point located? Show your work

If you used a converging lens with a focal length of 17 cm, where would you put an object that has a height of 5.5 cm to create a virtual image with a height of 13.8 cm? The image is located 22 cm away from the mirror. Show your work

Describe virtual images and how the simulation showed why they are virtual

A) The image is shrunken from 20 cm to 10cm.

B) The focal length is 19.7cm.

C) The object should be placed at 8.7 cm from the mirror.

Step-by-step explanation:

A) object distance(u) = 20cm

Focal length(f) = 7 cm

Image distance = v

(1/f) = (1/v) + (1/u)

(1/7) = (1/v) + (1/20)

(1/7) - (1/20) = (1/v)

(1/v) = (20-7) /140

(1/v) = 14/140

= 1/10

V = 10 cm

The image is shrunken from 20 cm to 10cm

B) image distance (v) = 51.5cm

Object distance (u) = 32cm

Focal length  = f

(1/f) = (1/v) + (1/u)

= (1/51.5) + (1/32)

= (32+51.5) /1648

= (83.5) /1648

= 1/19.7

f = 19.7 cm

The focal length is 19.7cm

C) focal length (f) = 17cm

Object height = 5.5cm

Image height = 13.8cm

Image distance (v) = 22cm

M = -v/u = 13.8/5.5

-22/u = 13.8/5.5

u =( 22 x 5.5) /13.8

u = 8.7 cm

The object should be placed at 8.7 cm from the mirror

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet