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Q. An astronomical telescope has an angular magnification of magnitude $5$ for distant objects. The separation between the objective and the eye-piece is $36 \,cm$ and the final image is formed at infinity. The focal length $ {{f}_{o}} $ of the objective and $ {{f}_{e}} $ of the eye-piece are respectively

KEAMKEAM 2010Ray Optics and Optical Instruments

Solution:

$ \frac{{{f}_{o}}}{{{f}_{e}}}=5 $
$ {{f}_{o}}=5{{f}_{e}} $
$ {{f}_{o}}+{{f}_{e}}=36 $
$ 5{{f}_{e}}=6\,cm $ and $ {{f}_{e}}=30\,cm $