InterviewSolution
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InterviewSolution
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Explain the construction of refracting telescope by figure and obtain the equation of its magnification. |
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Answer» Solution :To observe very huge celestial bodies an Astronomical Telescope is used. Its ray diagram is shown in the figure. In this telescope two convex lenses are kept in such a way that their principal axis coincide. The lens facing the object is called objective and the lens near the eye is known as eye-piece. Here the diameter and the focal length of the objective are greater than that of the eyepiece.  The cyc-piccc can move to & fro in the telescope tube when the telescope is focussed on a distant object, parallel rays coming from this object form a real, inverted and small image AB on the second principal FOCUS of the objective. This image is th object for the eye-piece. Eye-piece is moved to & fro to get the final an MAGNIFIED inverted image A.B. of the origin object at a certain distance. In such telescope, rays from the object ai refracted by the objective and the image FORMED. Such telescope is called refractin telescope. Figure of Astronomical Telescope is show above. Magnification of the telescope, `m=("angle subtended by the final image with eye")/("angle subtended by the object with the objective or eye")` `therefore m= (beta)/(alpha)` But TAN`beta=(h)/(f_e)` and for small angle `tan beta~~beat` `therefore beta=(h)/(f_e)` thus tan`alpha=(h)/(f_0)` and for small angle tan`alpha~~alpha` `therefore alpha=(h)/(f_0)` `therefore` Telescope magnification `m=(beta)/(alpha)=(h//f_e)/(h//f_0)` `therefore m=(f_0)/(f_e)` This equation shows that to increase the magnification of the telescope, focal length of the objective should be increased and focal length of the eye-piece should be reduced. `f_0+f_e` is the optical length of the telescope. So, length of the tube `L ge f_0+f_e` For a telescope, light gathering power and resolving power are very important. |
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