aChordate said:Homework Statement
An object is 12.0 cm in front of a convex mirror. When the convex mirror is replaced
with a flat mirror, the image distance is 5.0 cm farther away from the mirror. What is the
focal length of the convex mirror?
Homework Equations
f=-1/2R
1/do+1/di=1/f
The Attempt at a Solution
(1/0.12m)+1/di=1/f
?
12+5cm=17cm=0.17m
(1/0.17m)+1/di=1/f
?
ehild said:How is the image distance related to the object distance in case of a flat mirror?
ehild
The image is at 7 cm distance from the mirror, but the mirror is convex, is the image virtual or real?aChordate said:di=12cm for the flat mirror
di=7cm for the convex mirror?
aChordate said:Homework Statement
An object is 12.0 cm in front of a convex mirror. When the convex mirror is replaced
with a flat mirror, the image distance is 5.0 cm farther away from the mirror. What is the
focal length of the convex mirror?
Homework Equations
f=-1/2R
1/do+1/di=1/f
The Attempt at a Solution
(1/0.12m)+1/di=1/f
?
12+5cm=17cm=0.17m
(1/0.17m)+1/di=1/f
?
The focal length of a convex mirror is the distance from the mirror's surface to the point where all reflected light rays converge. It is denoted by the letter "f" and is measured in meters.
The focal length of a convex mirror can be determined by using the formula f = R/2, where R is the radius of curvature of the mirror. The radius of curvature is the distance from the center of the mirror to its curved surface.
The focal length of a convex mirror is inversely proportional to the size of the image formed. This means that as the focal length increases, the size of the image decreases, and vice versa.
Yes, the focal length of a convex mirror can be negative. This occurs when the object is placed between the focal point and the mirror's surface. In this case, the image formed is virtual and upright.
The focal length of a convex mirror has a direct impact on the magnification of the image formed. As the focal length decreases, the magnification increases, resulting in a larger and closer image. Conversely, as the focal length increases, the magnification decreases, resulting in a smaller and farther image.