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(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48xx10^(6)m, and the radius of lunar orbit is 3.8xx10^(8)m. |
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Answer» Solution :(a) Here, `f_(0) = 15 m` and `f_(e) =1.0 cm = 10^(-2) m` `THEREFORE` ANGULAR magnification `|m| =f_(0)/f_(e) =15/10^(-2) = 1500` (b) Diameter of MOON `D = 3.48 xx 10^6` m and distance of moon from earth = radius of lunar orbit of moon r = `3.8 xx 10^8 m`. The moon SUBTENDS an angle a at the telescope objective where `alpha = D/r`. The objective lens forms image of moon in its focal plane. If SIZE of image be h, then `alpha =h/f_(0)` `rArr h/f_(0) =D/r, therefore h=D/r f_(0)) =(3.48 xx 10^(6) xx 15)/(3.8 xx 10^(8) = 13.7 xx 10^(-7) m` or `13.7` cm |
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