Question

3.
An object with height ho = 4.8 mm is placed upright a distance s0 = 16 cm from the center of a concave mirror. The image is located a distance si = 7.0 cm from the center of the mirror.
(a) What is the focal length of the mirror?
(b) What is the size and orientation of the image?
(c) Where should the object be placed so that the image size is the same as the objectsize?
(d) What are the image distance and the magnification when the object is placed at a distance s0 = 4.1 cm from the center of the mirror?

Answer 1

@theEric

Answer 2

@Festinger

Answer 3

@Preetha @radar

Answer 4

@Fields

Answer 5

@Fifciol

Answer 6

@chenping

Answer 7

http://www.physicsclassroom.com/Class/refln/u13l3f.cfm
hope this does help :)

Answer 8

If i get answers to it,, it would be nice, plz

Answer 9

I will leave the geometry used to derive this equation out and jump directly to the formula:

\[\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\]
Where p is the object distance
q is the image distance and f is the focal point.

For spherical mirror sign convention p,q and R are all positive in AIR space.

Taking the values:
\[\frac{1}{f}= \frac{1}{16}+\frac{1}{7}\]
solve for f to get the focal length, which is 4.87cm.

For magnification:
\[M=\frac{h'}{h}=\frac{-q}{p}\]
Put in the relevant values for q, p and h to find h', the size of the image. The minus sign out mean it is inverted.

For part (c), solve for M=-1=-q/p. You should see that q=p.

part (d):
\[\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\]
p=s0=4.1. f=4.87,
q=-25.9cm

minus here means the image is behind the mirror, aka virtual!

\[M=\frac{25.9}{4.1}=6.3\times magnification\]

Answer 10

hank you :D

Answer 11

thank* you