MHT CET · Physics · Ray Optics
A convex lens of focal length ' \(\mathrm{f}\) ' produces a real image ' \(n\) ' time the size of the object. The image distance is
- A \(\mathrm{f}(\mathrm{n}+1)\)
- B \(f(n-1)\)
- C \(\frac{\mathrm{f}}{(\mathrm{N}+1)}\)
- D \(\frac{\mathrm{f}}{(\mathrm{N}-\mathrm{l})}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{f}(\mathrm{n}+1)\)
Step-by-step Solution
Detailed explanation
The image is real and hence inverted.
\(
\therefore \frac{\mathrm{v}}{\mathrm{u}}=-\mathrm{n} \text { or } \mathrm{u}=-\frac{\mathrm{v}}{\mathrm{n}}
\)
By lens equation, \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\(
\begin{aligned}
& \frac{1}{v}+\frac{n}{v}=\frac{1}{f} \\
& v=f(1+n)
\end{aligned}
\)
\(
\therefore \frac{\mathrm{v}}{\mathrm{u}}=-\mathrm{n} \text { or } \mathrm{u}=-\frac{\mathrm{v}}{\mathrm{n}}
\)
By lens equation, \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\(
\begin{aligned}
& \frac{1}{v}+\frac{n}{v}=\frac{1}{f} \\
& v=f(1+n)
\end{aligned}
\)
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