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NEET PHYSICSEasy

A small telescope has an objective of focal length 140 cm140\text{ cm} and an eyepiece of focal length 5.0 cm5.0\text{ cm}. The magnifying power of the telescope for viewing a distant object is:

A

28

B

17

C

32

D

34

Step-by-Step Solution

  1. Magnifying Power Formula: For an astronomical telescope in normal adjustment (viewing a distant object with the final image formed at infinity), the magnifying power (mm) is given by the ratio of the focal length of the objective lens (fof_o) to the focal length of the eyepiece (fef_e). m=fofem = \frac{f_o}{f_e}
  2. Given Data: Focal length of objective, fo=140 cmf_o = 140\text{ cm} Focal length of eyepiece, fe=5.0 cmf_e = 5.0\text{ cm}
  3. Calculation: m=1405.0=28m = \frac{140}{5.0} = 28
  4. Conclusion: The magnifying power of the telescope is 28.
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