MHT CET · Physics · Ray Optics
A convex lens \(\mathrm{T}\) is used to form an image whose size is one fourth that of size of object. Then the object distance is
- A \(2\mathrm{f}\)
- B \(5 \mathrm{f}\)
- C \(4 \mathrm{f}\)
- D \(3 \mathrm{f}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{f}\)
Step-by-step Solution
Detailed explanation
Magnification, \(\mathrm{m}=-\frac{1}{4}\)
(Since the image is real and inverted, it is taken as negative)
\(
\begin{aligned}
& \therefore \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=-\frac{1}{4} \\
& \mathrm{v}=-\frac{\mathrm{u}}{4}
\end{aligned}
\)
Lens formula, \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\(
\begin{aligned}
& -\frac{4}{u}-\frac{1}{u}=\frac{1}{f} \\
& u=-5 f
\end{aligned}
\)
(Since the image is real and inverted, it is taken as negative)
\(
\begin{aligned}
& \therefore \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=-\frac{1}{4} \\
& \mathrm{v}=-\frac{\mathrm{u}}{4}
\end{aligned}
\)
Lens formula, \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\(
\begin{aligned}
& -\frac{4}{u}-\frac{1}{u}=\frac{1}{f} \\
& u=-5 f
\end{aligned}
\)
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