The image of an object formed by a mirror is real inverted and is of magnification -1 where is the object placed
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Answered by
2
At center of curvature.....
Answered by
2
Answer:
+ 30 cm
Explanation:
Given :
Magnification = - 1
Image distance = 30 cm
We know
m = - v / u
- 1 = 30 / u
u = - 30 cm
Now we have ,
1 / f = 1 / v + 1 / u
1 / f = 1 / - 30 - 1 / 30
f = - 15 cm
When object moved 20 cm
u' = - 30 + 20 = - 10 cm
We have f = - 15 cm
We have to find new image distance v'
Again ,
1 / f = 1 / v' + 1 / u'
- 1 / 15 = 1 / v' - 1 / 10
1 / v' = 1 / 30
v' = 30.
Nature of image as :
Virtual
Erect
Magnified.
Attachments:
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