A concave shaving mirror has a radius of curvature of 28.0 cm. For each of the following cases, find the magnification and determine whether the image formed is real or virtual and upright or inverted.

(a) an upright pencil placed 41.6 cm from the mirror.

(b) an upright pencil placed 28.0 cm from the mirror.

(c) an upright pencil placed 7.6 cm from the mirror.

To find the magnification and determine the nature (real or virtual) and orientation (upright or inverted) of the image formed by a concave shaving mirror, we can use the mirror formula:

1/f = 1/v - 1/u

where f is the focal length of the mirror, v is the image distance, and u is the object distance.

First, let's find the focal length of the mirror using the radius of curvature (R):

f = R/2 = 28.0 cm / 2 = 14.0 cm

Now, let's calculate the magnification (m) for each case using the formula:

m = -v/u

(a) For the upright pencil placed 41.6 cm from the mirror:
- Object distance (u) = 41.6 cm
- Image distance (v) = ?
Using the mirror formula:
1/14.0 = 1/v - 1/41.6
Solve for v:
1/v = 1/14.0 + 1/41.6
v = 17.9 cm
Now, calculate the magnification:
m = -17.9 / 41.6 = -0.43

The magnification is -0.43, which means the image is inverted.
Since the magnification is less than 1, the image formed is smaller than the object.
Therefore, the image is real and inverted.

(b) For the upright pencil placed 28.0 cm from the mirror:
- Object distance (u) = 28.0 cm
- Image distance (v) = ?
Using the mirror formula:
1/14.0 = 1/v - 1/28.0
Solve for v:
1/v = 1/14.0 + 1/28.0
v = 9.33 cm
Now, calculate the magnification:
m = -9.33 / 28.0 = -0.33

The magnification is -0.33, which means the image is inverted.
Since the magnification is less than 1, the image formed is smaller than the object.
Therefore, the image is real and inverted.

(c) For the upright pencil placed 7.6 cm from the mirror:
- Object distance (u) = 7.6 cm
- Image distance (v) = ?
Using the mirror formula:
1/14.0 = 1/v - 1/7.6
Solve for v:
1/v = 1/14.0 + 1/7.6
v = -5.38 cm (Negative sign indicates a virtual image)
Now, calculate the magnification:
m = -(-5.38) / 7.6 = 0.71

The magnification is +0.71, which means the image is upright.
Since the magnification is greater than 1, the image formed is larger than the object.
Therefore, the image is virtual and upright.