The image formed by a spherical mirror is real inverted and is magnification -2 if the image is a distance of 30 from the mirror where is the object placed find focal length of the mirror
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m= -v/u
-2 = -v/u
2= v/u
2u=v
but v = -30
thus u = -15
-2 = -v/u
2= v/u
2u=v
but v = -30
thus u = -15
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Focal length ??
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Answer:
Given :
Magnification m = - 2
Image distance v = 30 cm
We know :
m = - v / u
- 2 = - (- 30 ) / u
u = - 15 cm
We have mirror formula :
1 / f = 1 / v + 1 / u
1 / f = 1 / - 15 - 1 / 30
1 / f = - 1 / 10
f = - 10 cm
When the object is moved 10 cm towards the mirror.
Then : u' = u - 10 cm
= > u' = - 5 cm
Again putting in formula to get v'
1 / - 10 + 1 / 5 = 1 / v'
1 / v' = 2 - 1 / 10
v' = 10 cm
Therefore , the image is located 10 cm behind the mirror.
Also new magnification :
m' = -v' / u'
m' = - 10 / -5
m' = 2
Since magnification is positive the image is erect and virtual.
Also image is magnified in nature.
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