The resolving power of a microscope is increased by: MCQ with Answer and Explanation

The resolving power of a microscope is increased by:
A. Using oil immersion
B. Using light of longer wavelength
C. Decreasing the numerical aperture
D. Reducing the magnification
Answer: Option A
Solution (By JKSSB Mock Tests)
Resolving power (minimum resolvable distance) d = λ/(2NA), where NA is numerical aperture. Oil immersion increases NA by matching refractive index between slide and objective, reducing light refraction and increasing light collection. Longer wavelength (A) decreases resolving power; decreasing NA (C) worsens resolution; magnification (D) doesn't affect fundamental resolution limit. Memory aid: 'Higher NA or shorter λ ⇒ better resolution'. This application question tests optical instrument knowledge, common in competitive exams. Always link resolution to wave nature of light (diffraction limit) and instrument design factors.

Discuss this Question (0)

No comments yet. Be the first to start the discussion!

Practice More Physics Questions

Question #1
A physical quantity is measured as 12.345 ± 0.005 units. The relative error in this measurement is approximately:
A. 0.4%
B. 4%
C. 0.04%
D. 0.004%

Correct Answer: Option C


Explanation:
Relative error = (Absolute error) / (Measured value) = 0.005 / 12.345 ≈ 0.000405. Converting to percentage: 0.000405 × 100% ≈ 0.0405% ≈ 0.04%. Relative error quantifies precision independent of measurement scale. Absolute error (±0.005) indicates instrument precision. This concept is vital for comparing measurement reliability across different magnitudes. Memory tip: Percentage error = (Absolute error / True value) × 100. Competitive exams often test this calculation with varying significant figures to assess numerical proficiency.

This question belongs to: Science Physics
Question #2
The unit of pressure in SI is
A. Torr
B. Atmosphere
C. Bar
D. Pascal

Correct Answer: Option D


Explanation:
Pascal = N/m². Named after Blaise Pascal. Bar = 10⁵ Pa, atmosphere = 101325 Pa, torr = 133.3 Pa. SI unit is Pascal.

This question belongs to: Science Physics
Question #3
The unit of electric potential is
A. Coulomb
B. Ohm
C. Volt
D. Ampere

Correct Answer: Option C


Explanation:
Volt = J/C. Named after Alessandro Volta. Potential difference drives current. 1 V = 1 J of work per coulomb charge moved.

This question belongs to: Science Physics