MHT CET · Physics · Ray Optics
A convex lens of focal length ' \(f\) ' \(m\) forms a real, inverted image twice in size of the object. The object distance from the lens in metre is
- A 0.5 f
- B 0.66 f
- C f
- D 1.5 f
Answer & Solution
Correct Answer
(D) 1.5 f
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Focal length }=\mathrm{f} \\
& \mathrm{~m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-2 ...(given)\\
\therefore \quad & \mathrm{v}=-2 \mathrm{u}
\end{aligned}\)
From lens formula,
\(\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{-2 u}-\frac{1}{u}=\frac{1}{f} \\ & \frac{u+2 u}{2 u^2}=\frac{1}{f} \\ & \frac{3 u}{2 u^2}=\frac{1}{f} \\ & \frac{1.5}{u}=\frac{1}{f} \\ & \therefore \quad u=1.5 \mathrm{f}\end{aligned}\)
& \text { Focal length }=\mathrm{f} \\
& \mathrm{~m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-2 ...(given)\\
\therefore \quad & \mathrm{v}=-2 \mathrm{u}
\end{aligned}\)
From lens formula,
\(\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{-2 u}-\frac{1}{u}=\frac{1}{f} \\ & \frac{u+2 u}{2 u^2}=\frac{1}{f} \\ & \frac{3 u}{2 u^2}=\frac{1}{f} \\ & \frac{1.5}{u}=\frac{1}{f} \\ & \therefore \quad u=1.5 \mathrm{f}\end{aligned}\)
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