Question

# A point object $\mathrm{O}$ is placed on the principal axis of a convex lens of focal length $\mathrm{f}=20\mathrm{cm}$ at a distance of $40\mathrm{cm}$ to the left of it. The diameter of the lens is $10\mathrm{cm}$. An eye is placed $60\mathrm{cm}$ to right of the lens and a distance $\mathrm{h}$ below the principal axis. The maximum value of $\mathrm{h}$ to see the image is____.

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Solution

## Step 1. Given dataFocal length $\mathrm{f}=20\mathrm{cm}$Distance $=40\mathrm{cm}$We have to find the maximum value of distance $\mathrm{h}$.Step 2. Formula used.A lens is a piece of a refracting medium bounded by two surfaces, at least one of which is a curved surface.We will calculate the image distance by using the lens formula.This formula represents the relation between object distance $\mathrm{u}$, image distance $\mathrm{v}$, and focal length $\mathrm{f}$.$⇒$ $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$ Here, $\mathrm{f}=20\mathrm{cm}$According to the sign of convention, object distance $\mathrm{u}=-40\mathrm{cm}$ $\mathrm{v}=?$Step 3. Calculate the image distance.By using the lens formula, we get,$\frac{1}{20}=\frac{1}{\mathrm{v}}-\frac{1}{-40}$$\frac{1}{20}=\frac{1}{\mathrm{v}}+\frac{1}{40}$ $\frac{1}{\mathrm{v}}=\frac{1}{20}-\frac{1}{40}$$\frac{1}{\mathrm{v}}=\frac{1}{40}$$\mathrm{v}=40\mathrm{cm}$So, the image distance is $40\mathrm{cm}$, which means the image is formed at the center of curvature.Draw the ray diagramStep 4. Find the value of distance.According to the above figure, $∆\mathrm{CED}$ and $∆\mathrm{CAB}$ are in symmetric form.So,$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{DC}}$$\frac{5}{\mathrm{h}}=\frac{40}{20}$$\frac{5}{\mathrm{h}}=2$$\mathrm{h}=\frac{5}{2}$$\mathrm{h}=2.5\mathrm{cm}$The maximum value of $\mathrm{h}$ to see the image is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$.

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