Ray Optics

JEE Main level, single correct, medium

207 of 295

Question 207 of 295

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\mathrm{cm}. The area of the image is 9 times that of the square. The focal length of the lens is :

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Answer: (D) 30 cm30 \mathrm{~cm}

If side of object square =ℓ=\ell and side of image square =ℓ′=\ell^{\prime}
From question, ℓ′2ℓ=9\frac{\ell^{\prime 2}}{\ell}=9
or ℓ′ℓ=3\frac{\ell^{\prime}}{\ell}=3
i.e., magnification m=3\mathrm{m}=3
u=−40 cm\mathrm{u}=-40 \mathrm{~cm}
v=3×40=120 cm\mathrm{v}=3 \times 40=120 \mathrm{~cm}
f=?\mathrm{f}=?
From formula, 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
1120−1−40=1f or, 1f=1120+140=1+3120∴f=30 cm\begin{aligned} & \frac{1}{120}-\frac{1}{-40}=\frac{1}{\mathrm{f}} \\ & \text { or, } \frac{1}{\mathrm{f}}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120} \therefore \mathrm{f}=30 \mathrm{~cm} \end{aligned}

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