Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
of the lens. The focal length and type of the lens should be
Solution:
1172 Upvotes
Verified Answer
The correct answer is:
\(+18.75 \mathrm{~cm} \) and convex lens
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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.