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A convex lens produces an image of a real object on a screen with a magnification of \frac{1}{2}. When the lens is moved \mathrm{30\ cm} away from the object, the magnification of the image on the screen is \mathrm{2}. The focal length of the lens is:

Option: 1

\mathrm{30\ cm}


Option: 2

\mathrm{60\ cm}


Option: 3

\mathrm{20\ cm}


Option: 4

\mathrm{15\ cm}


Answers (1)

best_answer

In the first case, let \mathrm{u=-x}
Then, \mathrm{v=+\frac{x}{2}}
\mathrm{\therefore \frac{2}{x}+\frac{1}{x}=\frac{1}{f} \quad \text { or } \quad \frac{3}{x}=\frac{1}{f}...........(i)}
In the second case, \mathrm{u=-(x-30) \Rightarrow v=2(x-30)}
\mathrm{\therefore \frac{1}{2(\mathrm{x}-30)}+\frac{1}{\mathrm{x}-30}=\frac{1}{\mathrm{f}}=\frac{3}{\mathrm{x}}\ \ .........(ii)}
On solving eqs. \mathrm{(i)} and \mathrm{(ii)}, we get 
\mathrm{f=20\ cm}

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himanshu.meshram

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