the focal length of this mirror is 6 cm and the object is located 8 cm away from the mirror. calculate the position of the image formed by the mirror. suppose the converging mirror is replaced by a diverging mirror with the same radius of curvature that is the same distance from the object for this mirror how does the size of the image compare with that of the object? is it larger than the object, smaller than the object, or the same size as the object?

1. For your first question, use the lens equation

1/do + 1/di = 1/f, where f is the focal length and do is the object distance, 8 cm.
2. For your second question, replace f = 6 cm with f = -6 cm, keep do = 8 cm and solve for di. The magnification is the absolute value of di/do

28

To calculate the position of the image formed by the mirror, we can use the mirror equation:

1/f = 1/d_o + 1/d_i

where:
f = focal length of the mirror
d_o = object distance from the mirror
d_i = image distance from the mirror

Given:
f = 6 cm
d_o = 8 cm

Plugging these values into the mirror equation, we can solve for d_i:

1/6 = 1/8 + 1/d_i
1/6 - 1/8 = 1/d_i
(8 - 6) / (8 * 6) = 1/d_i
2/48 = 1/d_i
1/24 = 1/d_i
d_i = 24 cm

Therefore, the position of the image formed by the mirror is 24 cm away from the mirror.

If the converging mirror is replaced by a diverging mirror with the same radius of curvature and the same distance from the object, the image formed would be virtual and located on the same side as the object.

Regarding the size of the image compared to the object, a diverging mirror always produces virtual, upright, and reduced images. Thus, the size of the image formed by the diverging mirror will be smaller than the size of the object.

To calculate the position of the image formed by the mirror, we can use the mirror formula:

1/f = 1/v - 1/u

Where,
f = focal length of the mirror
v = distance of the image from the mirror
u = distance of the object from the mirror

Given:
Focal length (f) = 6 cm
Object distance (u) = 8 cm

Substituting these values into the formula:

1/6 = 1/v - 1/8

To solve this equation, we can cross-multiply:

8v - 6u = uv
8v = uv + 6u
v(8 - u) = 6u
v = (6u) / (8 - u)

Now, let's calculate the position of the image.

v = (6 * 8) / (8 - 6)
v = 48 / 2
v = 24 cm

Therefore, the position of the image formed by the mirror is 24 cm away from the mirror surface.

Now, let's move on to the next part of your question where we replace the converging mirror with a diverging mirror, but with the same radius of curvature and distance from the object.

For a diverging mirror, the image formed is always virtual and located on the same side as the object. The image is formed by extending back the reflected rays.

Since the image distance (v) is positive for virtual images, the image distance will be negative in this case.

So, the position of the virtual image formed by the diverging mirror will be -24 cm (24 cm on the same side as the object but in the opposite direction).

Now, to determine the size of the image compared to the object, we can use the magnification formula:

magnification (m) = -v/u

Given:
Object distance (u) = 8 cm
Image distance (v) = -24 cm

Substituting these values into the formula:

m = -(-24) / 8
m = 24 / 8
m = 3

Therefore, the magnification of the image formed by the diverging mirror is 3. This means that the size of the image is 3 times larger than the size of the object.