the converse mirror used to view has radius of curvature 6 metre and object located in 12 metre from the mirror find the position nature and size of the image
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Explanation:
Given,
Radius of curvature, R= 3 m
Object distance, u= -5 m
Let the image distance= v
and height of image = h1
therefore,
Focal length, f =R/3 = 3/2 = 1.5 m
Using lens formula,
1/v +1/u = 1/f
1/v= 1/f-1/u
= 1/1.5 -(-1/5) =0.867m
so, v= 1.15 m
The image is formed at a distance of 1.15 m behind the mirror.
Now, magnification,m = image distance/object distance = -v/u
= -1.15/(-5) = 0.23
So, the image is virtual and erect.
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