A distant hot air balloon subtends 0.25^o at | vedclass

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(C) The angular magnification $M$ of an astronomical telescope in normal adjustment (relaxed eye) is given by $M = \frac{\beta}{\alpha} = \frac{f_0}{f

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$5$

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(C) The angular magnification $M$ of an astronomical telescope in normal adjustment (relaxed eye) is given by $M = \frac{\beta}{\alpha} = \frac{f_0}{f_e}$,where $\beta$ is the angle subtended by the image at the eyepiece and $\alpha$ is the angle subtended by the object at the objective lens.Given $\alpha = 0.25^o$ and $\beta = 1.5^o$,we have $M = \frac{1.5}{0.25} = 6$.Thus,$\frac{f_0}{f_e} = 6$,which implies $f_0 = 6f_e$.The length of the telescope tube $L$ is given by $L = f_0 + f_e = 35 \ cm$.Substituting $f_0 = 6f_e$ into the length equation: $6f_e + f_e = 35 \ cm$.$7f_e = 35 \ cm$.$f_e = 5 \ cm$.

$A$ distant hot air balloon subtends $0.25^o$ at the objective lens of an astronomical telescope. The image subtends $1.5^o$ at the eyepiece when viewed with a relaxed eye. The objective and eyepiece are in a $35 \ cm$ long tube. Find the focal length of the eyepiece lens in $cm$.

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