A man at the back of the crowd watches a parade by holding a plane mirror just above his head. The parade passes 6m behind his head and the mirror is 0.25 in front of the man. How far does the image in the mirror appear to be from the man
To find out how far the image in the mirror appears to be from the man, we need to understand the concept of virtual images formed by mirrors.
In this case, the man is holding a plane mirror just above his head. When we look at an object in a mirror, we see its virtual image. The distance of the virtual image from the mirror is equal to the distance of the object from the mirror.
Given:
Distance from the parade to the man's head (object distance) = 6m
Distance from the man's head to the mirror (mirror distance) = 0.25m
Since the man is holding the mirror just above his head, we can assume that the observer's eyes and the mirror form a right angle triangle along with the object (the parade) and the virtual image.
Using the Pythagorean theorem, we can find the distance of the virtual image from the man:
(6m)^2 + (0.25m)^2 = (distance of the virtual image)^2
Simplifying the equation:
36m^2 + 0.0625m^2 = (distance of the virtual image)^2
36.0625m^2 = (distance of the virtual image)^2
Taking the square root of both sides:
√(36.0625m^2) = distance of the virtual image
Distance of the virtual image ≈ 6.01m
Therefore, the distance of the image in the mirror appears to be approximately 6.01 meters from the man.
To find the distance the image in the mirror appears to be from the man, we can use the concept of virtual image formation in plane mirrors.
Let's assume the distance from the man to the mirror is represented by 'd1', and the distance from the mirror to the virtual image is represented by 'd2'.
Given:
Distance of the parade behind the man = 6m
Distance of the mirror in front of the man = 0.25m
Using the mirror formula, we have:
1/d1 + 1/d2 = 1/f
Since the mirror is planar (flat), the focal length (f) is considered to be infinite. Therefore, we can simplify the equation to:
1/d1 + 1/d2 = 0
Using the given values, we can plug them into the equation:
1/0.25 + 1/d2 = 0
Now let's solve for d2:
1/0.25 + 1/d2 = 0
4 + 1/d2 = 0
1/d2 = -4
d2 = -1/4 (m)
Since distance cannot be negative, this means there is no real image formed in the mirror. However, there is a virtual image that appears to be -1/4 meters from the man. In other words, the image appears to be 0.25 meters in front of the mirror located at the same distance from the man.