Find the position, nature and size of the image formed by a convex lens of focal length 12 cm of
an object 5 cm high placed at a distance 20 cm from it.
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Using the lens law
1÷12-1÷20=1÷x
=10-120-6÷120
4÷120
1÷30
30=x
thus the image is 30 cm behind the lens.
thus the image is 30 cm behind the lens.the image is erect and virtual because the object is between the focus and centre of curvature.
thus the image is 30 cm behind the lens.the image is erect and virtual because the object is between the focus and centre of curvature.using the formula for magnification if image
30÷20=3÷2
the image is enlarged.
☆Hope it's help you☆
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