Question

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

Answer 1

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.

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