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In an astronomical telescope, focal length of eye piece is `5 cm` and focal length of objective is `75 cm`. The final image is formed at the least distance of distinct vision `(=25 cm)` from the eye. What is the magnifying power of the telescope ? |
Answer» Here, `f_e = 5 cm, f_0 = 75 cm, m = ?, d = 25 cm` From `m = -(f_0)/(f_e)(1+(f_e)/(d)) = -(75)/(5) (1+(5)/(25)) = -15 xx 1.2 = -18.0`. |
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