An image formed by a convex mirror ( f= 24 cm) has a magnification of 0.15. Which way and by how much should the object be moved to double the size of the image?

Anyone Can answer this Please

Use the mirror equation.

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/mireq.html

Thanks

To determine the direction and distance by which the object should be moved to double the size of the image formed by a convex mirror, we can use the magnification formula:

magnification (m) = -v/u

Where:
magnification (m) = 0.15 (given)
v = image distance (distance of the image from the mirror)
u = object distance (distance of the object from the mirror)

Since a convex mirror always forms a virtual, diminished, and erect image, the magnification is always negative. However, for convenience, we'll consider the absolute value of the magnification (0.15) and assign the negative sign later, indicating the virtual nature of the image.

We can rewrite the magnification formula as:

m = -1/v
v = -1/m

Given the magnification (m) is 0.15, we can calculate the image distance (v):

v = -1/0.15
v = -6.67 cm (approximately)

The negative sign indicates a virtual image formed by a convex mirror.

To double the size of the image, we need to calculate the new image distance (v2) while keeping the same magnification:

magnification (m2) = 2 * 0.15
m2 = 0.3

Using the magnification formula:

m2 = -v2/u
0.3 = -v2/u

Since the magnification is negative, we can rewrite the equation as:

0.3 = v2/u

Since we know v2 = -6.67 cm, we can substitute the values and solve for u:

0.3 = -6.67/u

Cross-multiplying:

u = -6.67/0.3
u = -22.23 cm (approximately)

Therefore, to double the size of the image formed by a convex mirror with a magnification of 0.15, the object should be moved 22.23 cm toward the mirror. The negative sign, as before, denotes the direction (towards the mirror).