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
B
x1+x22
C
x1x2
D
x1x2

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  $$f = \sqrt {{x_1}{x_2}}$$ .Physics

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