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A thin rod of length (f/3) is lying alon...

A thin rod of length `(f/3)` is lying along the principal axis of a concave
mirror of focal length f.Image is real , magnified and inverted and one of the end or rod coincides with its image itself. Find length of the image.

Text Solution

Verified by Experts

Image is real, magnified and inverted. So, the given rod lies between f and C.
Further , one end of the rod is coinciding with its image itself. Therefore ,it is lying at C. So, the
thin rod CR is kept as shown below.
` (##DCP_V05_C30_S01_018_S01.png" width="80%">
We have to apply mirror formula only for point R
` u=-(5f)/(3)`, focal length `=-f`
Using the mirror formula `(1/v)+(1/u)=(1/f).`
` We have, `(1/v)+(1/(-(5f)/(3)))=(1/-f).`
Solving this equation, we get `v=(-5f)/(2)` or `(-2.5f) `
So, the image of rod CR is C' R' as shown below.
` (##DCP_V05_C30_S01_018_S02.png" width="80%">
So, image length `=C'R' =0.5f or (f/2).`
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