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Question

With a concave mirror, an object is placed at a distance $${ x }_{ 1 }$$ from the principal focus, on the principal axis. The image is formed at a distance $${ x }_{ 2}$$ from the principal focus. The focal length of the mirror is 


A
x1x2
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B
x1+x22
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C
x1x2
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D
x1x2
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Solution

The correct option is D $$\sqrt { { x }_{ 1 }{ x }_{ 2 } } $$

Given,

The image distance from the focus is given as $${x_2}$$, and object distance from the focus is $${x_1}$$, hence

$$v = f - {x_2}$$ and $$v = f - {x_1}$$

Now using the mirror formula

$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \Rightarrow \dfrac{1}{f} = \dfrac{1}{{f - {x_2}}} + \dfrac{1}{{f - {x_1}}}$$

$$\dfrac{1}{f} = \dfrac{{f - {x_1} + f - {x_2}}}{{\left( {f - {x_1}} \right)\left( {f - {x_2}} \right)}}$$

$$f\left( {2f - {x_1} - {x_2}} \right) = \left( {f - {x_1}} \right)\left( {f - {x_2}} \right)$$

$$2{f^2} - f{x_1} - f{x_2} = {f^2} - f{x_1} - f{x_2} + {x_1}{x_2}$$

$${f^2} = {x_1}{x_2} \Rightarrow f = \sqrt {{x_1}{x_2}}$$

Therefore the focal length is  $$f = \sqrt {{x_1}{x_2}}$$ .


1996658_1470877_ans_43120ca8c19f4a3a894973d727e52ea0.png

Physics

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