MHT CET · Physics · Ray Optics
A convex lens of focal length ' \(f\) ' produces a real image whose size is ' \(n\) ' times the size of an object. The distance of the object from the lens is
- A \(\frac{\mathrm{n}+1}{\mathrm{nf}}\)
- B \(\mathrm{f}\left(1-\frac{1}{\mathrm{n}}\right)\)
- C \(\frac{\mathrm{nf}}{\mathrm{n}+1}\)
- D \(\mathrm{f}\left(1+\frac{1}{\mathrm{n}}\right)\)
Answer & Solution
Correct Answer
(D) \(\mathrm{f}\left(1+\frac{1}{\mathrm{n}}\right)\)
Step-by-step Solution
Detailed explanation
Given,
\(\therefore \quad\) Image distance, \(v=-n u\)
\(\begin{aligned}
\quad \frac{1}{\mathrm{f}} & =\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{-\mathrm{nu}}-\frac{1}{\mathrm{u}}=\frac{\mathrm{u}+\mathrm{nu}}{n u^2} \\
\therefore \quad \frac{1}{\mathrm{f}} & =\frac{1+\mathrm{n}}{\mathrm{nu}} \\
\mathrm{u} & =\mathrm{f}\left(\frac{1+\mathrm{n}}{\mathrm{n}}\right)=\mathrm{f}\left(1+\frac{1}{\mathrm{n}}\right)
\end{aligned}\)
\(\therefore \quad\) Image distance, \(v=-n u\)
\(\begin{aligned}
\quad \frac{1}{\mathrm{f}} & =\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{-\mathrm{nu}}-\frac{1}{\mathrm{u}}=\frac{\mathrm{u}+\mathrm{nu}}{n u^2} \\
\therefore \quad \frac{1}{\mathrm{f}} & =\frac{1+\mathrm{n}}{\mathrm{nu}} \\
\mathrm{u} & =\mathrm{f}\left(\frac{1+\mathrm{n}}{\mathrm{n}}\right)=\mathrm{f}\left(1+\frac{1}{\mathrm{n}}\right)
\end{aligned}\)
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