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Standard XII

Physics

Image in Thin Lens Using Formula

When an small...

Question

When an small object is placed at a distance $x_{1}$ and $x_{2}$ from a lens on its principal axis in the same side, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then, the focal length of the lens is

x_{1} - x_{2}

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\frac{x_{1} - x_{2}}{2}

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\frac{x_{1} + x_{2}}{2}

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x_{1} + x_{2}

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Solution

The correct option is C $\frac{x_{1} + x_{2}}{2}$
$m = \frac{f}{f + u} m_{1} = \frac{f}{f - x_{1}} m_{2} = \frac{f}{f - x_{2}} m_{1} = - m_{2} 2 f = (x_{1} + x_{2}) f = \frac{x_{1} + x_{2}}{2}$

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When an small object is placed at a distance x1 and x2 from a lens on its principal axis in the same side, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then, the focal length of the lens is

The correct option is C x1+x22m=ff+um1=ff−x1m2=ff−x2m1=−m22f=(x1+x2)f=x1+x22

When an small object is placed at a distance $x_{1}$ and $x_{2}$ from a lens on its principal axis in the same side, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then, the focal length of the lens is

The correct option is C $\frac{x_{1} + x_{2}}{2}$
$m = \frac{f}{f + u} m_{1} = \frac{f}{f - x_{1}} m_{2} = \frac{f}{f - x_{2}} m_{1} = - m_{2} 2 f = (x_{1} + x_{2}) f = \frac{x_{1} + x_{2}}{2}$