The image of an object formed by a mirror is real, inverted and its magnification is -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 obtained?
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Answers
Answer:
It is stated that,
The magnificant of the image(m) = -1
Hence, it is a real image as we know that, if the value is obtained is negative then the formed image is real.
The distance of the image(v) = 30 cm
Let, the distance of the object be, 'u'
Fro the relationship we get that
m = v/u
⇒ -1 = 30/u
⇒ -u = 30
⇒ u = -30 m
Hence, the object is placed at a distance of 30 m beyond the centre of curvature.
Now,
If the object is moved 20 cm towards the mirror then, the distance of the object
= (-30 -20)
= - 50
= 50 m beyond the centre of curvature
Hence, the distance of the formed mirror will also be decreased by the equal length
So, the distance of the formed image = (30-20) m
= 10 m
So, magnificant of the image = (10/-50)
= -1/5
= -0.2
Hence, the formed image is real, inverted and smaller than the object.
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.
