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An object 5 cm in length is held 25 cm a...

An object `5 cm` in length is held `25 cm` away from a converging lens of focal length `10 cm`. Then the position and height of the image is

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Given, height of object = 5 cm
Position of object, u = -25 cm
Focal length of the lens, f = 10 cm
Hence, position of image, v =?
We know that,
`1/v-1/u=1/f`
`1/v+5=(1)/(10)`
So, `1/v=(1)/(10)-5`
So, `1/v=((5-2))/(50)`
That is, `1/v=(3)/(50)`
So, `v=(50)/(3)` = 16.66 cm
Thus, distance of image is 16.66 cm on the opposite side of lens.
Now, magnification = v/u
That is, m= 16.66/-25 = -0.66
Also, m= height of image/height of object
Or, -0.66 = height of image/5 cm
Therefore, height of image = -3.3 cm
The negative sign of height of image shows that an inverted image is formed.
Thus, position of image = At 16.66 cm on opposite side of lens
Size of image = -3.3 cm at the opposite side of lens

Nature of image: Real and inverted.
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