In a Galilean telescope,if the powers of an o | vedclass

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(C) The focal length of the objective lens is $f_o = \frac{1}{P_o} = \frac{1}{1.25} = 0.8 \, m = 80 \, cm$.The focal length of the eye lens is $f_e =

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$75 \, cm$ and $16$

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(C) The focal length of the objective lens is $f_o = \frac{1}{P_o} = \frac{1}{1.25} = 0.8 \, m = 80 \, cm$.The focal length of the eye lens is $f_e = \frac{1}{P_e} = \frac{1}{-20} = -0.05 \, m = -5 \, cm$.For relaxed vision (normal adjustment),the length of the Galilean telescope is given by $L = |f_o| - |f_e| = 80 \, cm - 5 \, cm = 75 \, cm$.The magnification of the Galilean telescope is given by $m = \frac{f_o}{|f_e|} = \frac{0.8}{0.05} = 16$.Thus,the length is $75 \, cm$ and the magnification is $16$.

In a Galilean telescope,if the powers of an objective and eye lens are respectively $+1.25 \, D$ and $-20 \, D$,then for relaxed vision,the length and magnification will be

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