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 30cm from the mirror ,where is the object placed ? Find the position of the image if the object is now moved 20cm towards the mirror .what is the nature of the image obrained? Justify your answer with the help of Ray diagram
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Given that image formed is realinverted and it's magnification is -1
If magnification is -1 then it means the nature of the image is real invertedand of the same size.
In this case the object is placed at C.
If the image is at a distance of 30 cm from the mirror then the object is also at a distance of 30 cm .
If the object is moved 20 cm towards the mirror then the position of the object is between the pole and the principal focus.
(focus=15)
when the object is moved 20 cm towards the mirror then the nature of the image is virtual erect and enlarged.
If magnification is -1 then it means the nature of the image is real invertedand of the same size.
In this case the object is placed at C.
If the image is at a distance of 30 cm from the mirror then the object is also at a distance of 30 cm .
If the object is moved 20 cm towards the mirror then the position of the object is between the pole and the principal focus.
(focus=15)
when the object is moved 20 cm towards the mirror then the nature of the image is virtual erect and enlarged.
Answered by
0
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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