KCET · Physics · Ray Optics
A candle placed \( 25 \mathrm{~cm} \) from a lens forms an image on a screen placed \( 75 \mathrm{~cm} \) on the other side
of the lens. The focal length and type of the lens should be
- A \(+18.75 \mathrm{~cm} \) and convex lens
- B \( -18.75 \mathrm{~cm} \) and concave lens
- C \( +20.25 \mathrm{~cm} \) and convex lens
- D \( -20.25 \mathrm{~cm} \) and concave lens
Answer & Solution
Correct Answer
(A) \(+18.75 \mathrm{~cm} \) and convex lens
Step-by-step Solution
Detailed explanation
Using lens formula, we have
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Here, \( v=75 \mathrm{~cm} ; \mathrm{u}=-25 \mathrm{~cm} \). Therefore
\[
\begin{array}{l}
\frac{1}{f}=\frac{1}{75}-\frac{1}{-25}=\frac{1}{75}+\frac{1}{25}=\frac{75+25}{75 \times 25}=\frac{100}{75 \times 25}=\frac{4}{75} \\
\Rightarrow f=\frac{75}{4}=18.75 \mathrm{~cm}
\end{array}
\]
Focal length of lens \( =+18.75 \mathrm{~cm} \)
Type of lens should be convex lens.
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Here, \( v=75 \mathrm{~cm} ; \mathrm{u}=-25 \mathrm{~cm} \). Therefore
\[
\begin{array}{l}
\frac{1}{f}=\frac{1}{75}-\frac{1}{-25}=\frac{1}{75}+\frac{1}{25}=\frac{75+25}{75 \times 25}=\frac{100}{75 \times 25}=\frac{4}{75} \\
\Rightarrow f=\frac{75}{4}=18.75 \mathrm{~cm}
\end{array}
\]
Focal length of lens \( =+18.75 \mathrm{~cm} \)
Type of lens should be convex lens.
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