NEETAceToday's NEET Daily

Microscopes and astronomical telescopes with magnifying powers: NEET MCQs with answers

49 questions on this topic of Optics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Optics freeAll Optics questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. Consider the following two statements regarding a compound microscope: Statement I: The intermediate image formed by the objective lens is real, inverted, and magnified. Statement II: The eyepiece acts as a simple magnifier to produce a final virtual, erect image relative to the intermediate image. Choose the correct option:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are true.

The objective lens forms a real, inverted, and magnified image between its center and the focal point of the eyepiece. The eyepiece then functions as a simple microscope, forming an erect, virtual, and magnified image with respect to this intermediate image (though inverted with respect to the original object). Both statements are entirely true.

medium

2. In a compound microscope, both the objective and the eyepiece have:

  1. A long-focus objective and short-focus eyepiece
  2. Equal and large focal lengths
  3. Short focal lengths, the objective being shorter
  4. Long focal lengths
Show answer and explanation

Answer: (C) Short focal lengths, the objective being shorter

High magnification m ≈ (L/fo)(D/fe) requires small fo and fe. The objective typically has the smaller focal length.

NCERT Class 12

hard

3. In a compound microscope, the focal lengths of objective and eyepiece are 1 cm and 2 cm respectively, and the tube length is 20 cm. If the final virtual image is formed at infinity, the distance of the object from the objective lens is:

  1. 1.025 cm
  2. 1.125 cm
  3. 1.200 cm
  4. 1.059 cm
Show answer and explanation

Answer: (D) 1.059 cm

When the final image is at infinity, the intermediate image is at the focal point of the eyepiece. Thus, v₀ = L - fₑ = 20 - 2 = 18 cm. Using 1/v₀ - 1/u₀ = 1/f₀: 1/18 - 1/u₀ = 1/1 => -1/u₀ = 1 - 1/18 = 17/18, giving |u₀| = 18/17 ≈ 1.059 cm.

medium

4. The magnifying power of a simple microscope (focal length f) when the image is at the near point D is:

  1. D - f
  2. 1 + D/f
  3. D/f - 1
  4. f/D
Show answer and explanation

Answer: (B) 1 + D/f

When the image is formed at the least distance of distinct vision D, the magnification of a simple magnifier is m = 1 + D/f. For the image at infinity it is D/f.

NCERT Class 12

easy

5. In a compound microscope, the tube length L is defined as the distance between:

  1. the objective lens and the eyepiece lens
  2. the object and the intermediate image
  3. the focus of the objective and the final virtual image
  4. the second focal point of the objective and the first focal point of the eyepiece
Show answer and explanation

Answer: (D) the second focal point of the objective and the first focal point of the eyepiece

According to NCERT, the tube length L of a compound microscope is the distance between the second focal point of the objective lens and the first focal point of the eyepiece. This is the length of the optical tube where light propagates between focal regions.

hard

6. In a compound microscope, the focal length of the objective is 1.2 cm and that of the eyepiece is 3.0 cm. An object is placed at 1.5 cm from the objective. If the final virtual image is formed at the least distance of distinct vision (D = 25 cm), the magnifying power of the microscope is:

  1. 37.3
  2. 25.2
  3. 45.5
  4. 18.7
Show answer and explanation

Answer: (A) 37.3

The magnification of the objective is m₀ = v₀ / |u₀|. Using 1/v₀ - 1/(-1.5) = 1/1.2, we get 1/v₀ = 1/1.2 - 1/1.5 = (5 - 4)/6 = 1/6, so v₀ = 6 cm. Thus, |m₀| = 6 / 1.5 = 4. The angular magnification of the eyepiece at D is mₑ = 1 + D/fₑ = 1 + 25/3 = 28/3 ≈ 9.33. The total magnifying power is m = |m₀| × mₑ = 4 × (28/3) = 112/3 ≈ 37.3.

medium

7. An astronomical telescope in normal adjustment has an angular magnification of 20 and the separation between its objective and eyepiece is 105 cm. The focal lengths of the objective and eyepiece are respectively:

  1. 95 cm and 10 cm
  2. 100 cm and 5 cm
  3. 90 cm and 15 cm
  4. 80 cm and 25 cm
Show answer and explanation

Answer: (B) 100 cm and 5 cm

In normal adjustment, m = f₀/f_e = 20, which gives f₀ = 20f_e. The length of the tube is L = f₀ + f_e = 21f_e = 105 cm. This yields f_e = 5 cm and f₀ = 20 × 5 = 100 cm.

medium

8. An astronomical telescope has an angular magnification of magnitude 12 when adjusted for infinity. If the objective lens has a power of +1.25 D, what is the power of the eyepiece lens?

  1. +15 D
  2. +10 D
  3. +12 D
  4. +5 D
Show answer and explanation

Answer: (A) +15 D

The focal length of the objective lens is f₀ = 100 / P₀ = 100 / 1.25 = 80 cm = 0.8 m. In normal adjustment, magnification m = f₀ / fₑ, so fₑ = f₀ / m = 80 cm / 12 = 20/3 cm = 0.2/3 m. The power of the eyepiece is Pₑ = 1 / fₑ = 3 / 0.2 = +15 D.

medium

9. An astronomical telescope has an objective of focal length 105 cm and an eyepiece of focal length 3.5 cm. If the final image is formed at the near point of distinct vision (D = 25 cm), what is the total magnifying power of the telescope?

  1. 34.2
  2. 30.0
  3. 25.8
  4. 42.0
Show answer and explanation

Answer: (A) 34.2

The magnifying power of an astronomical telescope when the final image is at the near point is given by m = (f₀ / fₑ) × (1 + fₑ / D). Here, f₀ = 105 cm, fₑ = 3.5 cm, and D = 25 cm. Thus, m = (105 / 3.5) × (1 + 3.5 / 25) = 30 × (1 + 0.14) = 30 × 1.14 = 34.2.

easy

10. Consider the following two statements regarding optical instruments: Statement I: In a compound microscope, both the objective lens and the eyepiece lens must have short focal lengths to achieve high magnification. Statement II: In an astronomical telescope, both the objective lens and the eyepiece lens must have large focal lengths to achieve high magnification. In light of the above statements, choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct: the magnification of a compound microscope is approximately m ≈ (L / f₀) × (D / fₑ), so both f₀ and fₑ should be small for large magnifying power. Statement II is incorrect: for an astronomical telescope, m = f₀ / fₑ, which requires a large focal length for the objective lens (f₀) but a small focal length for the eyepiece lens (fₑ).

medium

11. An astronomical telescope is used to view a 50 m tall tower situated at a distance of 2 km. The objective lens has a focal length of 100 cm and the eyepiece has a focal length of 2 cm. What is the height of the intermediate image of the tower formed by the objective lens?

  1. 1.0 cm
  2. 5.0 cm
  3. 2.5 cm
  4. 0.5 cm
Show answer and explanation

Answer: (C) 2.5 cm

The intermediate image is formed virtually in the focal plane of the objective because the object is far away (2000 m >> 1 m). The angle subtended by the tower at the objective is θ = height / distance = 50 / 2000 = 0.025 rad. Therefore, the height of the image formed by the objective is h' = f₀ × θ = 100 cm × 0.025 = 2.5 cm.

easy

12. A compound microscope uses an objective of focal length f₀ and an eyepiece of focal length fₑ. For viewing the final image at the near point D, the overall magnifying power m is given approximately by which expression?

  1. (f₀ / fₑ) × (1 + L / D)
  2. (f₀ / L) × (1 + D / fₑ)
  3. (L / f₀) × (D / fₑ - 1)
  4. (L / f₀) × (1 + D / fₑ)
Show answer and explanation

Answer: (D) (L / f₀) × (1 + D / fₑ)

The magnifying power of a compound microscope when the final image is at the near point D is the product of linear magnification of objective m₀ ≈ L/f₀ and angular magnification of eyepiece mₑ = (1 + D/fₑ). Thus, m ≈ (L / f₀)(1 + D / fₑ).

Keep going with Optics. All 503 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Optics