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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
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B
12f
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C
2f
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D
14f
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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, $$l_{img}=\dfrac{5f}{2}-\dfrac{4f}{2}$$

$$=\dfrac{f}{2}$$

67968_6053_ans_f6b8e4af9b484f1f8dd05802af718c74.png

Physics

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