Physics · Geometrical Optics
JEE Main 2024 — 1 February, Shift 1 — Question 49
The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm .
The focal length of the lens used is cm .
Answer: 15
Numerical answer — enter this value.
Step-by-step solution
\begin{array}{*{35}{r}}{} & \text{v}=3\text{u} \\{} & \text{v}-\text{u}=20\text{ }\!\!~\!\!\text{ cm} \\{} & 2\text{u}=20\text{ }\!\!~\!\!\text{ cm} \\{} & \text{u}=10\text{ }\!\!~\!\!\text{ cm} \\{} & \frac{1}{\left( -30 \right)}-\frac{1}{\left( -10 \right)}=\frac{1}{\text{f}} \\{} & \text{f}=15\text{ }\!\!~\!\!\text{ cm} \\\end{array}
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 1 February, Shift 1
- Subject
- Physics
- Chapter
- Geometrical Optics
- Topic
- Lenses and Their Combinations, Silvering of Lens