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  • A screen beaming a real image of magnification m1 formed by a convex lens is moved a distance . The ob

A screen beaming a real image of magnification m1 formed by a convex lens is moved a distance \mathrm{x}. The object is the moved until a new image of magnification \mathrm{m_{2}} is formed on the screen. The focal length of the lens is

Option: 1

\frac{x}{m_{2}-m_{1}}


Option: 2

\frac{x}{m_{1}-m_{2}}


Option: 3

\frac{x}{\sqrt{m_{1} m_{2}}}


Option: 4

None of these


Answers (1)

best_answer

\text{In first case, }\, \frac{1}{p}+\frac{1}{q}=\frac{1}{f}$ and $\frac{q}{p}=m_{1}
\mathrm{\Rightarrow \quad 1+m_{1}=\frac{q}{f} \quad \ldots(1)}

\mathrm{\text{In the second case }\frac{1}{q+x}+\frac{1}{p^{\prime}}=\frac{1}{f}}
\mathrm{And \, \frac{q+x}{p^{\prime}}=m_{2}}
\mathrm{\Rightarrow \quad m_{2}=\frac{q+x}{f}\quad \ldots(2)}
(1) and (2)
\mathrm{\Rightarrow \quad m_{2}-m_{1}=x / f \Rightarrow f=\frac{x}{m_{2}-m_{1}}}

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