The image of an object formed by a mirror is real and inverted and is of magnification 1. If the image is at a 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 obtained ? Justify your answer with the help of Ray diagram. "Please help me."
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Given the image formed is real inverted and its magnification is -1
if magnificaion 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 and image is formed at C.
if the image is at a distance of 30cm from the mirror then the object is also at a distance of 30cm from the mirror.
if the object is moved 20cm towards the mirror then the position of the object is between the pole and the principal focus (f=15cm)
When the object is moved 20cm towards the mirror then the nature of the image is virtual erect and enlarged
look the attachment for the diagram
if magnificaion 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 and image is formed at C.
if the image is at a distance of 30cm from the mirror then the object is also at a distance of 30cm from the mirror.
if the object is moved 20cm towards the mirror then the position of the object is between the pole and the principal focus (f=15cm)
When the object is moved 20cm towards the mirror then the nature of the image is virtual erect and enlarged
look the attachment for the diagram
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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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