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Difficulty: MediumThin Lenses, Optical Instruments, and Defects of Vision

An astronomical telescope in normal adjustment consists of an objective lens with a focal length of 60 cm60\text{ cm} and an eyepiece with a focal length of 5 cm5\text{ cm}. What is the angular magnification produced by the telescope and the separation distance between the two lenses?

  1. 1212 and 65 cm65\text{ cm}Answer
  2. B
    0.0830.083 and 65 cm65\text{ cm}
  3. C
    1212 and 55 cm55\text{ cm}
  4. D
    300300 and 65 cm65\text{ cm}

Answer

The angular magnification is 1212 and the separation distance between the lenses is 65 cm65\text{ cm}.
For an astronomical telescope in normal adjustment, the magnification is given by M=fofe=605=12M = \frac{f_o}{f_e} = \frac{60}{5} = 12, and the length of the telescope tube (distance between lenses) is L=fo+fe=60+5=65 cmL = f_o + f_e = 60 + 5 = 65\text{ cm}.

Step-by-Step Solution

1
Calculate the angular magnification (MM) of the telescope
M=fofe=605=12M = \frac{f_o}{f_e} = \frac{60}{5} = 12
In an astronomical telescope in normal adjustment, angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.
2
Calculate the separation distance (LL) between the two lenses
L=fo+fe=60+5=65 cmL = f_o + f_e = 60 + 5 = 65\text{ cm}
When in normal adjustment, the final image is formed at infinity, so the distance between the objective lens and eyepiece equals the sum of their focal lengths.

Key Concept

Astronomical telescope in normal adjustment (magnification and lens separation)
Estimated Time:1m 0s
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Topics

Thin Lenses, Optical Instruments, and Defects of Vision

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