Question
A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.
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