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Question

A rod of length $$10cm$$ lies along the principal axis of concave mirror of focal length $$10cm$$ in such a way that its end closer to the pole is $$20cm$$ away from the mirror. The length of the image is-


A
15cm
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B
2.5cm
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C
5cm
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D
10cm
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Solution

The correct option is C $$5cm$$
Rod $$AB$$ of length $$10 \ cm$$ lies on the axis in front of concave mirror such that end A lies at a distance of $$20 \ cm$$ from the mirror.
Using mirror formula :
$$\dfrac{1}{v}+\dfrac{1}{u} = \dfrac{1}{f}$$
Case A :  $$u = -20 \ cm$$     $$f = -10 \ cm$$
$$\therefore$$   $$\dfrac{1}{v_A}+\dfrac{1}{-20} = \dfrac{1}{-10}$$   
$$\implies \ v_A = -20 \ cm$$
Case B :  $$u = -30 \ cm$$     $$f = -10 \ cm$$
$$\therefore$$   $$\dfrac{1}{v_B}+\dfrac{1}{-30} = \dfrac{1}{-10}$$   
$$\implies \ v_B = -15 \ cm$$
Thus length of image  $$L = 20 - 15 =5 \ cm$$

Physics

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