KCET · Physics · Ray Optics
The size of the image of an object, which is at infinity, as formed by a convex lens of focal length \(30 \mathrm{~cm}\) is \(2 \mathrm{~cm}\). If a concave lens of focal length \(20 \mathrm{~cm}\) is placed between the convex lens and the image at a distance of \(26 \mathrm{~cm}\) from the lens, the new size of the image is
- A \(1.25 \mathrm{~cm}\)
- B \(2.5 \mathrm{~cm}\)
- C \(1.05 \mathrm{~cm}\)
- D \(2 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(2.5 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given, \(f_{1}=30 \mathrm{~cm}\),
\(\begin{aligned}
&f_{2}=-20 \mathrm{~cm}, h_{0}=2 \mathrm{~cm} \\
&u_{2}=(30-26)=4 \mathrm{~cm}
\end{aligned}\)
Using lens formula for concave lens,
\(\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}=-\frac{1}{20}+\frac{1}{4}=\frac{4}{20}=\frac{1}{5}\)
or \(v_{2}=5 \mathrm{~cm}\)
Magnification, \(m=\frac{v_{2}}{u_{2}}=\frac{5}{4}=1.25\)
Also, \(m=\frac{h_{i}}{h_{o}}=1.25\)
\(\therefore\) Size of new image,
\(h_{i}=h_{0} \times 1.25=2 \times 1.25=2.5 \mathrm{~cm}\)
\(\begin{aligned}
&f_{2}=-20 \mathrm{~cm}, h_{0}=2 \mathrm{~cm} \\
&u_{2}=(30-26)=4 \mathrm{~cm}
\end{aligned}\)
Using lens formula for concave lens,
\(\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}=-\frac{1}{20}+\frac{1}{4}=\frac{4}{20}=\frac{1}{5}\)
or \(v_{2}=5 \mathrm{~cm}\)
Magnification, \(m=\frac{v_{2}}{u_{2}}=\frac{5}{4}=1.25\)
Also, \(m=\frac{h_{i}}{h_{o}}=1.25\)
\(\therefore\) Size of new image,
\(h_{i}=h_{0} \times 1.25=2 \times 1.25=2.5 \mathrm{~cm}\)
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