There is a 12mm high object that is 10 cm from a concave mirror with a focal length of 17 cm. (a) Where (B) How high, and (C) what type type is its image.
For C i got real inverted and reduced but how do you do parts A and B
a) have you tried the mirror equation?
b) di/do=hi/ho
To determine the answers to parts A and B of the question, you can use the mirror equation and magnification formula. Here's how you can solve them:
(a) Where is the image located?
The mirror equation relates the object distance (u), image distance (v), and the focal length (f) of a concave mirror:
1/f = 1/v - 1/u
In this case, the object distance (u) is given as 10 cm, and the concave mirror's focal length (f) is given as 17 cm. We need to find the image distance (v). Rearranging the equation, we get:
1/v = 1/f - 1/u
1/v = 1/17 - 1/10
Now, we can calculate the image distance (v) by taking the reciprocal of the right-hand side of the equation:
v = 1 / (1/17 - 1/10)
v = 53.333 cm
Therefore, the image is located approximately 53.333 cm from the concave mirror.
(b) How high is the image?
The magnification formula gives the relationship between the height of the object (h) and the image (h'):
Magnification (m) = h'/h = -v/u
Here, we want to find the height of the image (h') relative to the height of the object (h). We already know the object distance (u) is 10 cm and image distance (v) is 53.333 cm.
m = -v/u
m = -53.333/10
m = -5.333
The negative sign indicates that the image is inverted. The magnification is 5.333.
Since we know the height of the object (12 mm), we can calculate the height of the image (h') using the magnification formula:
m = h'/h
5.333 = h'/12
To find h', we multiply both sides of the equation by 12:
h' = 5.333 * 12
h' = 64 mm
Therefore, the height of the image is 64 mm.
(c) The type of the image is determined by its orientation and size. In this case, since the magnification is positive, we would expect a real image. Since the image is also inverted (as determined in part C), the complete answer is that the image is real, inverted, and reduced.