Refraction and Mirrors, image position

[emg333]

Homework Statement

A concave mirror (R=54cm) faces a convex mirror of the same radius. The distance between the mirrors is 54cm, and their axes coincide. A candle is held between the mirrors, 10 cm from the convex mirror.
Consider light rays reflect off the concave mirror and then the convex.
Where do these rays form an image, in relation to the convex mirror?

Homework Equations

f= +/- .5R
1/s + 1/s' = 1/f

The Attempt at a Solution

i am not sure where to begin

 

[Chi Meson]

Have you drawn the situation yet?

The principle to keep in mind is that the image of the first mirror becomes the object of the second mirror. Solve the problem as a single concave mirror first, locate the image on your drawing, refigure the object distance to the second mirror, solve again.

 

[nlightner]

with this problem.

I would first start by understanding the concepts of refraction and mirrors. Refraction is the bending of light as it passes through different mediums, while mirrors are reflective surfaces that can change the direction of light. The distance between mirrors and their curvature can affect the position and size of the images formed.

In this problem, we have a concave mirror and a convex mirror facing each other with a distance of 54cm between them. The radius of both mirrors is also 54cm, meaning they have the same curvature. A candle is placed 10cm in front of the convex mirror.

To begin solving this problem, we can use the equations provided in the homework statement. The first equation, f= +/- .5R, tells us that the focal length (f) of a mirror is half of its radius (R). Since both mirrors have the same radius, they also have the same focal length.

Next, we can use the second equation, 1/s + 1/s' = 1/f, where s is the object distance, s' is the image distance, and f is the focal length. We know that the object distance is 10cm, and we want to find the image distance.

Plugging in the values, we get 1/10 + 1/s' = 1/27. Since 1/f is equal to 1/27, we can conclude that the image distance (s') is equal to 27cm. This means that the image formed by the convex mirror is 27cm away from the mirror.

Now, since the concave and convex mirrors are facing each other, the image formed by the convex mirror will act as the object for the concave mirror. Using the same equation, we can find the image distance for the concave mirror.

Plugging in the values, we get 1/27 + 1/s' = 1/27. This means that the image distance for the concave mirror is also 27cm. Therefore, the image formed by the concave mirror is also 27cm away from the mirror.

In conclusion, the rays of light that reflect off the concave mirror and then the convex mirror will form an image 27cm away from the convex mirror. This image will also act as the object for the concave mirror, forming another image 27cm away from the concave mirror.