the image of an object formed by a mirror is real inverted and is of magnification -1 if the image is at the distance of 30 CM from the mirror where is the object placed? find the position of the image if the object is now moved 20 CM towards the mirror what is the nature of the image of the image obtsined ? justify your answer with the help of a diagram.
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real and inverted
betwewn centre of curvature and focus
betwewn centre of curvature and focus
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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.
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