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In a compound microscope, the focal lengths of two lenses are $1.5 \mathrm{~cm}$ and $6.25 \mathrm{~cm}$. An object is placed at $2 \mathrm{~cm}$ from the objective and the final image is formed at $25 \mathrm{~cm}$ from the eye lens. The distance between the two lenses is .............. (in $\mathrm{cm}$ ).
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1169 Upvotes
Verified Answer
The correct answer is:
11
For, objective,
$$
u=-2 \mathrm{~cm}, f=1.5 \mathrm{~m}
$$
By $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we have,
$$
\frac{1}{v}=\frac{1}{1.5}-\frac{1}{2} v=6 \mathrm{~cm}
$$
For eyepiece,
$$
v=-25 \mathrm{~cm}, f=6.25 \mathrm{~cm}
$$
$\begin{aligned} \text { Using } \frac{1}{v}-\frac{1}{u} & =\frac{1}{f}, \text { we get } \\ \frac{1}{u} & =\frac{1}{-25}-\frac{1}{6.25} \\ u & =-5 \mathrm{~cm}\end{aligned}$
So, distance between two lenses = 11 cm.
$$
u=-2 \mathrm{~cm}, f=1.5 \mathrm{~m}
$$
By $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we have,
$$
\frac{1}{v}=\frac{1}{1.5}-\frac{1}{2} v=6 \mathrm{~cm}
$$
For eyepiece,
$$
v=-25 \mathrm{~cm}, f=6.25 \mathrm{~cm}
$$
$\begin{aligned} \text { Using } \frac{1}{v}-\frac{1}{u} & =\frac{1}{f}, \text { we get } \\ \frac{1}{u} & =\frac{1}{-25}-\frac{1}{6.25} \\ u & =-5 \mathrm{~cm}\end{aligned}$
So, distance between two lenses = 11 cm.
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