JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2025
- Easy
- MCQ
A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ________
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 15
Step-by-step Solution
Detailed explanation
Equivalent focal length
\(\begin{aligned}
& \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\
& =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\
& f=-60 \mathrm{~cm}
\end{aligned}\)
Lens formula
\(\begin{aligned}
& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
& \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\
& \mathrm{v}=-15 \mathrm{~cm}
\end{aligned}\)
\(\begin{aligned}
& \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\
& =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\
& f=-60 \mathrm{~cm}
\end{aligned}\)
Lens formula
\(\begin{aligned}
& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
& \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\
& \mathrm{v}=-15 \mathrm{~cm}
\end{aligned}\)
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