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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2013
- Hard
- Numerical
The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is \(40\,cm.\) The area of the image is \(9\,times\) that of the square. The focal length of the lens is........\(cm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(30\)
Step-by-step Solution
Detailed explanation
If side of object square \(=\ell\)
and side of image square \(=\ell^{\prime}\)
From question, \(\frac{\ell^{\prime 2}}{{\ell}^2 }=9\)
or \(\quad \frac{\ell^{\prime}}{\ell}=3\)
i.e., magnificationm \(=3\)
\(\mathrm{u}=-40 \mathrm{cm}\)
\(\mathrm{v}=3 \times 40=120\, \mathrm{cm}\)
\(f=?\)
From formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{120}-\frac{1}{-40}=\frac{1}{f}\)
or, \(\frac{1}{f}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120} \therefore f=30\, \mathrm{cm}\)
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