Question

You place a mirror 10 cm in front of your face, which creates an upright image that is twice the height of the object. You then move the mirror along the principle axis until it creates an inverted image twice the height of the object.

1. What kind of mirror must this be and why?

2. Where is the object with respect to the mirror’s focal length for each of the mirror

   positions? (ie. f < s < R)

3. Draw ray diagrams for the image created at each location of the mirror.

4. Considering the first location of the mirror, find the focal length and image position.

5. Considering the second location of the mirror, find the object and image positions.

6. How far did you move the mirror?

(I just need help making sure I drew the ray diagrams correctly)

Answer 1

see the ray diagrams :

first position :

second position :

a) For a real object, the ability to form both virtual image and real image depending on position of object, is done by concave mirror.

b) for first position :

u = object distance = -10 cm.

The object is at a distance less than the focal length from the mirror. [answer]

v1 = image distance = ?

f = focal length = ?

M = +2 (+ve for upright image).

M = -v1/u

=> v1 = -2u = +20 cm [answer]

image formed at 20 cm behind the mirror.

Applying mirroe equation for first position :

=> f = -20 cm. [answer]

focal length is 20 cm.

for second position :

let the mirror moved away by x cm.

u' = -(10+x) cm

v2 = ?

M = -2 (-ve for inverted image)

=> M = - v2/u'

=> v2 = 2u' = -2(10+x) cm

applying mirror equation :

=> x = 20 cm.

the object distance is 30 cm from the mirror. [answer]

the image formed is at 60 cm from the mirror. [answer]

the mirror is moved by 20 cm. [answer]