KCET · Physics · Ray Optics
The following figure shows a beam of light converging at point \(P\). When a concave lens of focal length \(16 \mathrm{~cm}\) is introduced in the path of the beam at a place shown by dotted line such that \(O P\) becomes the axis of the lens, the beam converges at a distance \(x\) from the lens. The value of \(x\) will be equal to
- A \(12 \mathrm{~cm}\)
- B \(24 \mathrm{~cm}\)
- C \(36 \mathrm{~cm}\)
- D \(48 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(48 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
As the beam of light is converging at point \(P\), so it will act as a virtual object for concave lens.
\(\therefore\) Objective distance, \(u=12 \mathrm{~cm}\)
Focal length of concave lens, \(f=-16 \mathrm{~cm}\)
Using lens formula,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\begin{array}{ll}\Rightarrow & \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48} \\ \text { or } & v=48 \mathrm{~cm} \\ \text { Given, } & v=x \mathrm{~cm} \\ \therefore & x=48 \mathrm{~cm}\end{array}\)
\(\therefore\) Objective distance, \(u=12 \mathrm{~cm}\)
Focal length of concave lens, \(f=-16 \mathrm{~cm}\)
Using lens formula,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\begin{array}{ll}\Rightarrow & \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48} \\ \text { or } & v=48 \mathrm{~cm} \\ \text { Given, } & v=x \mathrm{~cm} \\ \therefore & x=48 \mathrm{~cm}\end{array}\)
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