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Standard X

Physics

Lens Maker's Formula

The distance ...

Question

The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is

m

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\frac{1}{m}

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m + 1

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\frac{1}{m + 1}

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Solution

The correct option is D $\frac{1}{m + 1}$
Given, $u = - m f$
$∴ \frac{1}{v} - \frac{1}{(- m f)} = - \frac{1}{f}$
$\Rightarrow \frac{1}{v} = - \frac{1}{f} (1 + \frac{1}{m})$
$\Rightarrow - \frac{v}{f} = \frac{m}{m + 1}$
$\Rightarrow - \frac{v}{u} = (\frac{1}{1 + m})$

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The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is

The correct option is D $\frac{1}{m + 1}$
Given, $u = - m f$
$∴ \frac{1}{v} - \frac{1}{(- m f)} = - \frac{1}{f}$
$\Rightarrow \frac{1}{v} = - \frac{1}{f} (1 + \frac{1}{m})$
$\Rightarrow - \frac{v}{f} = \frac{m}{m + 1}$
$\Rightarrow - \frac{v}{u} = (\frac{1}{1 + m})$

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The distance between an object and a divergent lens is m times the focal length of the lens. The linear magnification produced by the lens is

The correct option is D 1m+1Given, u=−mf∴1v−1(−mf)=−1f⇒1v=−1f(1+1m)⇒−vf=mm+1⇒−vu=(11+m)

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The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is

The correct option is D $\frac{1}{m + 1}$
Given, $u = - m f$
$∴ \frac{1}{v} - \frac{1}{(- m f)} = - \frac{1}{f}$
$\Rightarrow \frac{1}{v} = - \frac{1}{f} (1 + \frac{1}{m})$
$\Rightarrow - \frac{v}{f} = \frac{m}{m + 1}$
$\Rightarrow - \frac{v}{u} = (\frac{1}{1 + m})$