3. Justify the fact that for a converging lens, more focal length ==> less light

Question

3. Justify the fact that for a converging lens, more focal length ==> less light-bending ability of the lens Hint: In the distant object method, you might remember fei (image distance). So more f ==> more i ==> more image distance. A lens with high light bending ability must form the image of a distant object closer to the lens (in other words, i should be small). Only a weak lens (with low light bending ability) should form an image far away from the lens (more i). See the diagram below to understand.

Explanation / Answer

First of all we have to understand that the convergin lens try to converge the light rays toward the focus point.
According to lens law if the rays are parallel to the principal axis then the lens converge them to focal point.
Now if a lens has more converging ability then then the image will be formed near to the pole or we can say focus is near to the pole that means i will be smaller.
But if the converging ability of the lens is less then image will be formed at larger distance which means the focal point that is "i" will be larger.
From the lens equation
1/di + 1/do = 1/f
If object is at infinity then 1/do = 0 , therefore
1/di = 1/f
di = f
hence nearer the image shorter the focal length or "i"
or more the ability to get converge.

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