An object in front of a concave mirror has a virtual image that is 16.5 cm from the mirror. The mirror's radius of curvature is 19.5 cm. (Include the sign of the value in your answers.)
(a) What is the object distance?
(b) What is the magnification?
What I find comfusing on these is to remember the sign conventions. What worked for me is to sketch a ray diagram first, get an order of magnitude answer from the ray diagram, then do the math to check signs.
To find the object distance and magnification in this scenario, we can use the mirror equation and magnification formula.
(a) To find the object distance, we can use the mirror equation:
1/f = 1/do + 1/di
where:
- f is the focal length of the mirror,
- do is the object distance, and
- di is the image distance.
In this case, we are given the radius of curvature (R) of the mirror, and we know that for a concave mirror, the focal length (f) is half the radius of curvature. Thus, f = R/2.
Given: R = 19.5 cm
So, the focal length, f = 19.5 cm / 2 = 9.75 cm.
We are also given that the image distance (di) is 16.5 cm. We need to find the object distance (do).
Plugging in these values into the mirror equation, we have:
1/9.75 = 1/do + 1/16.5
Rearranging the equation to solve for do, we get:
1/do = 1/9.75 - 1/16.5
Now, we can calculate do:
1/do = (16.5 - 9.75) / (9.75 * 16.5) = 0.069
do = 1 / 0.069 = 14.49 cm (the object distance)
Hence, the object distance is approximately 14.49 cm.
(b) To find the magnification, we can use the magnification formula:
magnification (m) = -di / do
Given that the image distance (di) is 16.5 cm and the object distance (do) is 14.49 cm, we can calculate the magnification:
m = -16.5 / 14.49 ≈ -1.14
Therefore, the magnification is approximately -1.14. The negative sign indicates that the image is inverted.