A telescope has an objective of focal length | vedclass

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(A) For the objective lens,the object distance $u_o = -300 \ cm$ and focal length $f_o = 100 \ cm$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u

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$-3$

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(A) For the objective lens,the object distance $u_o = -300 \ cm$ and focal length $f_o = 100 \ cm$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$,we get $\frac{1}{v_o} - \frac{1}{-300} = \frac{1}{100}$.This gives $\frac{1}{v_o} = \frac{1}{100} - \frac{1}{300} = \frac{2}{300}$,so $v_o = 150 \ cm$.The magnification of the objective is $m_o = \frac{v_o}{u_o} = \frac{150}{-300} = -0.5$.For the eyepiece,the image is formed at the least distance of distinct vision,so $v_e = -25 \ cm$. With $f_e = 5 \ cm$,the magnification of the eyepiece is $m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$.The total magnification is $M = m_o 	imes m_e = -0.5 	imes 6 = -3$.

$A$ telescope has an objective of focal length $100 \ cm$ and an eyepiece of focal length $5 \ cm$. The least distance of distinct vision is $25 \ cm$. The telescope is focused for distinct vision on a scale $3 \ m$ away from the objective. The magnification produced is . . . . . .

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