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NEET UG Optics — practice questions

Explore 3 free Optics questions in NEET UG Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Optics questions and exercises with answers

Showing questions 1–3 of 3.

Worked practice questions

A telescope has an objective lens of focal length $120\,\text{cm}$ and an eyepiece of focal length $5\,\text{cm}$. Its magnifying power is

  1. 12
  2. 20
  3. 24
  4. 125
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Correct answer: C. 24

For a telescope, $m=f_o/f_e$. Thus $m=120/5=24$. **So the telescope produces an angular magnification of 24.**

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The magnifying power of a telescope in normal adjustment is given by

  1. $m=\dfrac{f_e}{f_o}$
  2. $m=\dfrac{f_o}{f_e}$
  3. $m=1+\dfrac{D}{f_e}$
  4. $m=\dfrac{L}{f_o}\times\dfrac{D}{f_e}$
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Correct answer: B. $m=\dfrac{f_o}{f_e}$

For a telescope, the magnifying power is the ratio of the angle subtended by the image at the eye to that subtended by the object, and in normal adjustment it is $m=f_o/f_e$. Options C and D are microscope relations.

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The approximate magnifying power of a compound microscope is given by

  1. $m=\dfrac{f_o}{f_e}$
  2. $m=\dfrac{L}{f_e}\times\dfrac{D}{f_o}$
  3. $m=\dfrac{L}{f_o}\times\dfrac{D}{f_e}$
  4. $m=\dfrac{D}{L}\times\dfrac{f_o}{f_e}$
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Correct answer: C. $m=\dfrac{L}{f_o}\times\dfrac{D}{f_e}$

For a compound microscope, the approximate magnifying power is $m=(L/f_o)\times(D/f_e)$, where $L$ is the distance between the focal points of the objective and eyepiece, $f_o$ is the focal length of the objective, and $f_e$ is the focal length of the eyepiece. **Option A is the telescope formula, not the microscope formula.**

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