SINCE \[m=\frac{{{f}_{o}}}{{{f}_{e}}}\]                    Also \[m=\frac{ANGLE\ subtended\ by\ the\ IMAGE}{Angle\ subtended\ by\ the\ object}\]            \[\THEREFORE \frac{{{f}_{o}}}{{{f}_{e}}}=\frac{\ALPHA }{\beta }\]\[\Rightarrow \alpha =\frac{{{f}_{o}}\times \beta }{{{f}_{e}}}=\frac{60\times 2}{5}={{24}^{o}}\] "> SINCE \[m=\frac{{{f}_{o}}}{{{f}_{e}}}\]                    Also \[m=\frac{ANGLE\ subtended\ by\ the\ IMAGE}{Angle\ subtended\ by\ the\ object}\]            \[\THEREFORE \frac{{{f}_{o}}}{{{f}_{e}}}=\frac{\ALPHA }{\beta }\]\[\Rightarrow \alpha =\frac{{{f}_{o}}\times \beta }{{{f}_{e}}}=\frac{60\times 2}{5}={{24}^{o}}\] ">
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A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focussed on a distant object is such a way that parallel rays comes out from the eye lens. If the object subtends an angle 2o at the objective, the angular width of the image  [CPMT 1979; NCERT 1980;  MP PET 1992; JIPMER 1997; UPSEAT 2001]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 8 months ago

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SINCE \[m=\frac{{{f}_{o}}}{{{f}_{e}}}\]                    Also \[m=\frac{ANGLE\ subtended\ by\ the\ IMAGE}{Angle\ subtended\ by\ the\ object}\]            \[\THEREFORE \frac{{{f}_{o}}}{{{f}_{e}}}=\frac{\ALPHA }{\beta }\]\[\Rightarrow \alpha =\frac{{{f}_{o}}\times \beta }{{{f}_{e}}}=\frac{60\times 2}{5}={{24}^{o}}\]
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