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(a) Draw the ray diagram of an astronomical telescope when the final images is formed at infinity. Write the expression for the resolving power of the telescope. (b) An astronomical telescope has an objective lens of focal length 20m and eyepiece offocal length 1 cm. (i) Find the angular magnification of the telescope. (ii) If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is 3.5 xx 10^(6) m and the radius of lunar orbit is 3.8 xx 10^(8)m. |
Answer» Solution :(a) A ray digram of an astronomical telescope when the final image is formed at infinity is drawn here.  Resolving POWER of the telescope is given as: R.R. `=A/(1.22 lambda)` where A is the aperature (diameter) of the objective LENS of telescope and `lambda` is the wavelength of light coming from distant object. (b) Here `f_(0)=20cm, f_(e)=1cm=0.01m`, diameter of moon `D=3.5 xx 10^(6)m`, and distance of moon lunar orbbit `=d=3.8 xx 10^(8)m` (i) Angular agnification of telescope `=f_(0)/f_(e)=(20)/(0.01) =2000` (ii) Since image of moon is formed by the objective lens at its FOCAL PLANE. If diameter of image by `D_(V)`, then `D_(i)/f_(0)=D/d` `rArr D_(i)=(f_(0) xx D)/(d)=(20 xx 3.5 xx 10^(6))/(3.8 xx 10^(8)) =0.184 m=18.4m` |
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