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(C) For a convex lens,the object distance $u = -40 \, cm$ and focal length $f = +20 \, cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{

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$2.5$

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(C) For a convex lens,the object distance $u = -40 \, cm$ and focal length $f = +20 \, cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$. Thus,$v = 40 \, cm$. The image is formed at $40 \, cm$ to the right of the lens. The lens has a radius of $5 \, cm$. The ray passing through the top edge of the lens $(A)$ passes through the image point $(C)$ and reaches the eye at point $(E)$. From the similar triangles $\Delta ABC$ and $\Delta CDE$,we have $\frac{AB}{BC} = \frac{DE}{CD}$. Here,$AB = 5 \, cm$ (radius),$BC = 40 \, cm$ (image distance),$DE = h$,and $CD = 60 \, cm - 40 \, cm = 20 \, cm$. Therefore,$\frac{5}{40} = \frac{h}{20} \Rightarrow h = \frac{5 	imes 20}{40} = 2.5 \, cm$.

Similar Questions

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

The diameter of a plano-convex lens is $6 \ cm$ and its thickness at the centre is $3 \ mm$. If the speed of light in the material of the lens is $2 \times 10^8 \ m/s$,find the focal length of the lens in $cm$.

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