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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Ray Optics

BITSAT 2014 · Physics · Ray Optics

The magnifying power of a telescope is 9. When it is adjusted for...

The magnifying power of a telescope is 9.9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20cm20 \mathrm{ cm}. The focal length of lenses are

Options

  1. A18cm,2cm18 \mathrm{ cm}, 2 \mathrm{ cm}
  2. B11cm,9cm11 \mathrm{ cm}, 9 \mathrm{ cm}
  3. C10cm,10cm10 \mathrm{ cm}, 10 \mathrm{ cm}
  4. D15cm,5cm15 \mathrm{ cm}, 5 \mathrm{ cm}

Correct answer

A. 18cm,2cm18 \mathrm{ cm}, 2 \mathrm{ cm}

Step-by-step solution

f0fe=9,∴f0=9fe\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=9, \quad \therefore \mathrm{f}_{0}=9 \mathrm{f}_{\mathrm{e}} Also f0+fe=20(∵\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=20(\because final image is at infinity )9fe+fe=20,fe=2cm,∴f0=18cm)9 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=20, \mathrm{f}_{\mathrm{e}}=2 \mathrm{ cm}, \quad \therefore \mathrm{f}_{0}=18 \mathrm{ cm}
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