Question

# A thin rod of length $$\dfrac{f}{3}$$ lies along the axis of a concave mirror of focal length $$f$$. One end of its image touches at end of the rod. The length of the image is :

A
f
B
12f
C
2f
D
14f

Solution

## The correct option is B $$\dfrac{1}{2}f$$Since one end of rod touches image that end is at center of curvature.From lens formula, $$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$$$\dfrac{1}{v}-\dfrac{3}{5f}=\dfrac{-1}{f}$$$$\dfrac{1}{v}=\dfrac{-5}{5f}+\dfrac{3}{5f}$$$$\dfrac{1}{v}=\dfrac{-2}{5f}$$$$v=-\dfrac{5f}{2}$$Length of image, $$l_{img}=\dfrac{5f}{2}-\dfrac{4f}{2}$$$$=\dfrac{f}{2}$$Physics

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