How Far is the Second Object from the Convex Mirror?

[lizzyann31]

Homework Statement

An object is located 14 cm in front of a convex mirror, the image being 7 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?


I know I need to use the mirror equation... but I am totally LOST!

 

[rl.bhat]

First of all find the focal length of the convex mirror and magnification. In this problem magnification v/u = 1/2 = I/o. When the When the size of the object doubles keepin image height the same. magnification becomes 1/4.
So V'/U' = 1/4. Write V' in terms of U', substitute in the mirror formula to get U'

 

[nlightner]

I understand that the mirror equation is a fundamental principle in optics that relates the object distance, image distance, and focal length of a curved mirror. In this case, we are dealing with a convex mirror, which means the focal length is positive and the image formed will always be virtual and upright.

To solve for the location of the second object, we can use the mirror equation, which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

We know that the first object is located 14 cm in front of the convex mirror, so do = -14 cm (negative sign indicates it is located in front of the mirror). The image distance, di, is given as 7 cm behind the mirror, so di = 7 cm.

To find the focal length, we can use the fact that the image height is the same for both objects. This means that the magnification, M = hi/ho = -di/do, where hi is the image height and ho is the object height. Since the magnification is the same for both objects, we can set their ratios equal to each other and solve for the focal length.

For the first object: M = hi/ho = -di/do = hi/ho = -7/ho

For the second object: M = hi/ho = -di/do = hi/(2ho) = 7/(2ho)

Setting these two ratios equal to each other and solving for ho, we get ho = 2 cm.

Now, we can plug this value for ho into the mirror equation and solve for the object distance, do, for the second object.

1/f = 1/do + 1/di

1/f = 1/do + 1/7

Substituting ho = 2 cm and solving for do, we get do = 4 cm.

Therefore, the second object is located 4 cm in front of the convex mirror. This also makes sense intuitively, as the second object is twice as tall as the first one, so it would need to be closer to the mirror in order for the image to be the same height.