A 50kg person standing still in ice skates throws their helmet to the right at 25 m/s while on a frictionless surface. This give
1 answer:
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Answer:
1) When <
(hence
< f ) and they are both in front of the mirror (positive), the image will be larger and inverted
2) When >
(and
< f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted
3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror
Explanation:
The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation
Where:
f = Focal length of the mirror = R/2
= Image distance from the mirror
= Object distance from the mirror
= Image height
= Object height
is positive for an object placed in front of the mirror and negative for an object placed behind the mirror
is positive for an image formed in front of the mirror and negative for an image formed behind the mirror
m is positive when the orientation of the image and the object is the same
m is negative when the orientation of the image and the object is inverted
f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)
∴ When <
(hence
< f ) and they are both in front of the mirror (positive), the image will be larger and inverted
When >
(and
< f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted
When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.