A spherical glass ornament is 7.00 cm in diameter. An object is placed 11.8 cm away

from the ornament.
a) Where will its image form?

b) What is the magnification of the image

The focal length of the convex mirror surface is

f = -R/2 = -1.75 cm

For the image distance di, use
1/do + 1/di = 1/f = -1/1.75
1/di = -1/1.75 -1/11.8 = -0.6562
di = -1.524 cm (1.524 cm behind the bulb surface)
Magnification = |di/do| = 1.524/11.8
= 0.129

To answer these questions, we can use the thin lens equation, which is given by:

1/f = 1/di + 1/do

where:
- f is the focal length of the lens,
- di is the image distance from the lens,
- do is the object distance from the lens.

a) To determine where the image will form, we need to find the image distance (di).

Given:
- Diameter of the spherical glass ornament = 7.00 cm (which means the radius of the ornament is 3.50 cm).
- Object distance (do) = 11.8 cm.

First, we need to find the focal length (f) of the lens. Since the ornament is considered a spherical mirror, the focal length will be half of the radius of curvature (R) of the ornament.

The radius of the ornament is 3.50 cm, so the radius of curvature (R) is also 3.50 cm.

Therefore, the focal length (f) will be half of the radius of curvature:
f = R/2 = 3.50 cm / 2 = 1.75 cm.

Now, we can use the thin lens equation to find the image distance (di):

1/f = 1/di + 1/do

Substituting the values into the equation:
1/1.75 = 1/di + 1/11.8.

To find di, we rearrange the equation:
1/di = 1/1.75 - 1/11.8.

Calculating this gives us:
1/di = 0.5714 - 0.0847.

Simplifying:
1/di = 0.4867.

To find di, we take the reciprocal of both sides of the equation:
di = 1 / 0.4867.

Calculating this gives us:
di = 2.056 cm.

Therefore, the image will form at a distance of approximately 2.056 cm from the lens.

b) To find the magnification (m) of the image, we can use the following formula:

m = -di / do.

Using the values we already have:
di = 2.056 cm.
do = 11.8 cm.

Substituting these values into the formula:
m = -2.056 cm / 11.8 cm.

Calculating this gives us:
m ≈ -0.1746.

Therefore, the magnification of the image is approximately -0.1746. The negative sign indicates that the image is inverted.