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A small telescope has an objective lens of focal length `144 cm` and an eye-piece of focal length `6.0 cm`. What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ?A. `0.75 cm`B. `1.38 cm`C. `1.0 m`D. `1.5 m` |
Answer» Correct Answer - D The separation between the objective and the eyepiece = Length of the telescope tube `f = f_(0) + f_(e)` Here, `f_(0) = 144 cm, = 1.44 m, f_(e) = 6.0 cm = 0.06 m` `:. F = 1.44 + 0.06 = 1.5 m` |
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