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A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 $\mathrm{mm}$ apart. If the radius of curvature of the large mirror is $220 \mathrm{~mm}$ and the small mirror is $140 \mathrm{~mm}$, where will the final image of an object at infinity be?
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For objective mirror (concave), $u=-\infty$
Radius of curvature $R=-220 \mathrm{~mm}=-22 \mathrm{~cm}$
Focal length $f=\frac{R}{2}=-11 \mathrm{~cm}$
Image will be obtained at focus.
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}+\frac{1}{-\infty}=\frac{1}{-11}$ or $v=-11 \mathrm{~cm}$
Image formed by objective mirror act as a virtual object
for secondary mirror.
Distance of virtual object for secondary mirror.
$u_0=(11-2) \mathrm{cm}=+9 \mathrm{~cm}$
Focal length of convex mirror used $=14 \mathrm{~cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}=\frac{1}{14}-\frac{1}{9}$ or $v=-25.5 \mathrm{~cm}$
Radius of curvature $R=-220 \mathrm{~mm}=-22 \mathrm{~cm}$
Focal length $f=\frac{R}{2}=-11 \mathrm{~cm}$
Image will be obtained at focus.
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}+\frac{1}{-\infty}=\frac{1}{-11}$ or $v=-11 \mathrm{~cm}$
Image formed by objective mirror act as a virtual object
for secondary mirror.
Distance of virtual object for secondary mirror.
$u_0=(11-2) \mathrm{cm}=+9 \mathrm{~cm}$
Focal length of convex mirror used $=14 \mathrm{~cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}=\frac{1}{14}-\frac{1}{9}$ or $v=-25.5 \mathrm{~cm}$
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