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Draw a ray diagram for the formation of image of a distant object by an astronomical telescope in normal adjustment position. Deduce the expression for its magnifying power. |
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Answer» Solution :Magnifying power (or ANGULAR magnification) of an astronomical telescope is defined as the ratio of the angle SUBTENDED by the final image at eye (`beta`) to the angle subtended by object directly (`alpha`) at eye. As the object lies at very LARGE distance, angle subtended by the object at Q (the optical centre of objective) may be CONSIDERED as almost the same as the angle subtended by the object at the eye. Therefore, Angular magnification `m = beta/alpha` As angles `alpha`and `beta` , are small, therefore, `alpha = tanalpha` and `beta = tan beta` In `triangleA.B.C_(2)`, `tanbeta_(2)(A.B.)/(C_(2)B.)` and in `triangleA.B.C_(1), tan alpha =(A.B.)/(C_(1)B.)` `m=-f_(0)/f_(e)` Here -ve sign indicates that the final image is an inverted image. From the above relation it is clear that for higher magnifying power, the focal length `f_0` of the objective lens should be as high as possible and the focal length `f_e` of the eyepiece should be least possible. |
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