An object is placed 15cm in front of a concave mirror of focal length 10cm
Find:the position
Nature
Magnification of image formed
Given:
Object distance (u) = -15cm
Focal length (f) = -10cm (since concave mirror)
Using the mirror formula:
1/f = 1/v + 1/u
Substitute the known values:
1/(-10) = 1/v + 1/(-15)
-1/10 = 1/v - 1/15
-3/30 = 1/v
-1/10 = 1/v
Therefore, v = -10cm
The negative sign indicates that the image is virtual and formed on the same side as the object. So, the image distance is -10cm.
To find the magnification, we use the formula:
magnification (m) = -v/u
Substitute the known values:
m = -(-10)/-15
m = 10/15
m = 2/3
The positive magnification value indicates that the image is erect.
Therefore, the image is formed at 10cm in front of the mirror, it is virtual and erect, and the magnification of the image is 2/3.