A man standing 1.25 m in front of a shaving mirror produces an inverted image 25.0 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the actual size of his chin?
This question was added to a previous thread, and has already been answered there.
To find out how close the man should stand to form an upright image of his chin that is twice the actual size, we can use the mirror formula:
1/f = 1/di + 1/do
where:
- f is the focal length of the mirror
- di is the distance of the image from the mirror (positive if the image is real and negative if the image is virtual)
- do is the distance of the object from the mirror (positive if the object is in front of the mirror)
From the given information, we know that the image formed by the shaving mirror is inverted and located 25.0 cm in front of the mirror, so di = -25.0 cm = -0.25 m. We also know that the man is standing 1.25 m in front of the mirror, so do = 1.25 m.
We can rearrange the formula to solve for the focal length:
1/f = 1/di + 1/do
1/f = 1/-0.25 + 1/1.25
1/f = -4 + 0.8
1/f = -3.2
f = -1/3.2 = -0.3125
Since the focal length of the mirror is negative, we know it is a concave mirror. Now, we can use the magnification formula to find out how close the man should stand to form an upright image that is twice the actual size:
magnification = -di/do
Given that the magnification of the image should be 2 (twice the actual size), we can substitute the known values into the equation:
2 = -(-0.25)/1.25
2 = 0.25/1.25
2 = 1/5
Now, let's solve the equation to find the distance of the object from the mirror (do):
2 = 1/5
10 = 1/do
do = 1/10
do = 0.1 m
Therefore, the man should stand approximately 0.1 meters or 10 centimeters away from the shaving mirror in order to form an upright image of his chin that is twice the actual size.