An image formed by a convex mirror (f = -24.0 cm) has a magnification of 0.150. Which way and by how much should the object be moved to double the size of the image?

M= -q/p we want magnification doubled so M=.3 and -.3p = q

1/p + 1/q = 1/f
1/p + 1/-.3p = 1/-24
Solve for p

If you want to know the original p just do the same thing with M = .15

To double the size of the image formed by a convex mirror, we need to find out the new object distance. We can use the magnification formula to solve for this.

The formula for magnification (M) is:

M = -image distance / object distance

Given that the magnification (M) is 0.150 and the focal length (f) is -24.0 cm, we can plug these values into the formula and solve for the original image distance (di) in terms of the original object distance (do):

0.150 = -di / do

Now, let's calculate the original image distance (di) in terms of the original object distance (do):

di = -0.150 * do

To double the size of the image, we need to calculate the new object distance (do').

The formula for magnification (M') for the new object distance is:

M' = -image distance' / object distance'

Since we want to double the size of the image, the new magnification (M') will be 2 times the original magnification (M):

M' = 2 * M = 2 * 0.150 = 0.300

Now we can plug the new magnification (M') and the focal length (f) into the magnification formula and solve for the new object distance (do'):

0.300 = -image distance' / do'

Since we know that the image distance' is equal to the original image distance plus the change in position (delta di), we can rewrite the formula as:

0.300 = -(di + delta di) / do'

Rearranging the formula, we can solve for the new object distance (do'):

do' = -(di + delta di) / 0.300

Substituting the original image distance (di = -0.150 * do) into the equation above:

do' = -(-0.150 * do + delta di) / 0.300

We can solve this equation for the change in position (delta di):

delta di = -0.300 * do' + 0.150 * do

Now we can substitute the given focal length (f = -24.0 cm) into the equation for delta di:

delta di = -0.300 * do' + 0.150 * do
-24 = -0.300 * do' + 0.150 * (-do)

Now we can solve for the new object distance (do'):

-24 = -0.300 * do' - 0.150 * do

Rearranging the formula, we get:

0.300 * do' = -24 + 0.150 * do

Simplifying the equation:

0.300 * do' = 0.150 * do - 24

Now we can solve for the new object distance (do') by dividing both sides of the equation by 0.300:

do' = (0.150 * do - 24) / 0.300

Calculating this, we find:

do' = (0.150 * do - 24) / 0.300

Therefore, the object needs to be moved by a distance of (0.150 * do - 24) / 0.300 to double the size of the image formed by the convex mirror.

To determine the displacement required to double the size of the image formed by a convex mirror, we can use the mirror formula:

1/f = 1/v - 1/u

Where:
f is the focal length of the convex mirror (given as f = -24.0 cm)
v is the image distance
u is the object distance

First, let's calculate the initial image distance using the given magnification:

m = -v/u

Where:
m is the magnification (given as 0.150)

0.150 = -v/u

Rearranging the equation, we can find "v":

v = -0.150u

Now, substitute the value of "v" in the mirror formula and solve for "u":

1/(-24.0 cm) = 1/(-0.150u) - 1/u

Simplifying the equation:

-1/24.0 = -1/u + 1/(-0.150u)

To combine the fractions, let's find a common denominator of 120u:

-5u/120u = -120u/120u + 8u/120u

Simplifying further:

-5u/120u = (-120u + 8u)/120u

-5u/120u = -112u/120u

Now, equating the numerators:

-5u = -112u

Solving for "u":

-112u + 5u = 0

-107u = 0

u = 0

From the equation, we find that the object distance "u" is zero. This result implies that the object is located at the focal point of the convex mirror.

Since there is no object distance in this case, we conclude that moving the object closer or farther away from the convex mirror will not allow us to double the size of the image.