Resolving Power & Optical Instruments — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Resolving Power & Optical Instruments MCQs with step-by-step solutions (20 questions). Part of Wave Optics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Resolving Power & Optical Instruments · medium · theory
Rayleigh's criterion states that two point objects are just resolved when the central maximum of one diffraction pattern falls on:
A. The edge of the aperture
B. The first minimum of the other ✓ Correct
C. The central maximum of the other
D. The second maximum of the other
Solution: At this separation the combined intensity shows a shallow dip between the two peaks, which the eye can just detect.
Q2 — Resolving Power & Optical Instruments · medium · theory
The resolving power of a telescope of objective diameter $D$ for light of wavelength $\lambda$ is:
A. $1.22\lambda D$
B. $\dfrac{\lambda}{D}$
C. $\dfrac{1.22\lambda}{D}$
D. $\dfrac{D}{1.22\lambda}$ ✓ Correct
Solution: Resolving power is the reciprocal of the limit of resolution, so a bigger objective resolves finer detail.
Q3 — Resolving Power & Optical Instruments · easy · theory
The limit of resolution of a telescope having objective diameter $D$ is:
A. $\dfrac{D}{1.22\lambda}$
B. $\dfrac{2.44\lambda}{D}$
C. $\dfrac{1.22\lambda}{D}$ ✓ Correct
D. $\dfrac{0.61\lambda}{D}$
Solution: This is Airy's result for the smallest angular separation that a circular aperture can resolve.
Q4 — Resolving Power & Optical Instruments · easy · theory
The resolving power of a microscope increases when the wavelength of light used is:
A. Kept constant
B. Increased
C. Decreased ✓ Correct
D. Made zero
Solution: The limit of resolution is proportional to $\lambda$, so shorter wavelengths reveal finer detail — the reason ultraviolet microscopes exist.
Q5 — Resolving Power & Optical Instruments · easy · theory
To improve the resolving power of a telescope, one should:
A. Use longer wavelength light
B. Decrease the diameter of the objective
C. Increase the focal length of the eyepiece only
D. Increase the diameter of the objective ✓ Correct
Solution: A larger aperture reduces the diffraction spread of each star image, which is why research telescopes have very large mirrors.
Q6 — Resolving Power & Optical Instruments · medium · theory
An electron microscope achieves far higher resolving power than an optical microscope because electrons have:
A. A much longer wavelength
B. No electric charge
C. A much larger mass
D. A much shorter de Broglie wavelength ✓ Correct
Solution: Accelerated electrons have wavelengths thousands of times shorter than visible light, and the limit of resolution scales with wavelength.
Q7 — Resolving Power & Optical Instruments · medium · theory
Two stars are said to be just resolved by a telescope when their angular separation is:
A. Zero
B. Equal to the limit of resolution of the telescope ✓ Correct
C. Much smaller than the limit of resolution
D. Equal to the diameter of the objective
Solution: Below this angle the two diffraction discs merge into one and the pair appears as a single star.
Q8 — Resolving Power & Optical Instruments · hard · theory
The resolving power of a diffraction grating increases with:
A. The total number of illuminated lines and the order of the spectrum ✓ Correct
B. An increase in the wavelength only
C. A decrease in the number of lines
D. A decrease in the order of the spectrum
Solution: The grating resolving power is $nN$, so more lines and higher orders both sharpen the spectral separation.
Q9 — Resolving Power & Optical Instruments · medium · numerical
A telescope has an objective of diameter $1\text{ m}$ and is used with light of $500\text{ nm}$. Its limit of resolution is:
A. $6.1 \times 10^{-5}\text{ rad}$
B. $1.22 \times 10^{-6}\text{ rad}$
C. $5 \times 10^{-7}\text{ rad}$
D. $6.1 \times 10^{-7}\text{ rad}$ ✓ Correct
Solution: $\Delta\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22 \times 5 \times 10^{-7}}{1} = 6.1 \times 10^{-7}\text{ rad}$.
Q10 — Resolving Power & Optical Instruments · medium · numerical
A telescope objective of diameter $0.1\text{ m}$ is used with light of $600\text{ nm}$. Its limit of resolution is:
A. $6 \times 10^{-6}\text{ rad}$
B. $7.32 \times 10^{-7}\text{ rad}$
C. $1.22 \times 10^{-5}\text{ rad}$
D. $7.32 \times 10^{-6}\text{ rad}$ ✓ Correct
Solution: $\Delta\theta = \dfrac{1.22 \times 6 \times 10^{-7}}{0.1} = 7.32 \times 10^{-6}\text{ rad}$.
Q11 — Resolving Power & Optical Instruments · hard · numerical
A telescope of objective diameter $2\text{ m}$ is used with light of $500\text{ nm}$. Its resolving power is approximately:
A. $3.05 \times 10^{-7}$
B. $1.64 \times 10^6$
C. $6.1 \times 10^7$
D. $3.28 \times 10^6$ ✓ Correct
Solution: Resolving power $= \dfrac{D}{1.22\lambda} = \dfrac{2}{1.22 \times 5 \times 10^{-7}} \approx 3.28 \times 10^6$.
Q12 — Resolving Power & Optical Instruments · easy · numerical
If the wavelength of light used in a telescope is halved, its resolving power:
A. Becomes four times
B. Doubles ✓ Correct
C. Halves
D. Remains unchanged
Solution: Resolving power $\propto \dfrac{1}{\lambda}$, so halving the wavelength doubles it.
Q13 — Resolving Power & Optical Instruments · easy · numerical
If the diameter of a telescope objective is doubled, its limit of resolution:
A. Doubles
B. Halves ✓ Correct
C. Becomes four times
D. Remains unchanged
Solution: Limit of resolution $= \dfrac{1.22\lambda}{D} \propto \dfrac{1}{D}$, so a bigger aperture resolves a smaller angle.
Q14 — Resolving Power & Optical Instruments · hard · numerical
A telescope has an objective of diameter $5\text{ m}$ and works at $550\text{ nm}$. Its limit of resolution is approximately:
A. $6.71 \times 10^{-7}\text{ rad}$
B. $1.1 \times 10^{-7}\text{ rad}$
C. $1.34 \times 10^{-6}\text{ rad}$
D. $1.34 \times 10^{-7}\text{ rad}$ ✓ Correct
Solution: $\Delta\theta = \dfrac{1.22 \times 5.5 \times 10^{-7}}{5} \approx 1.34 \times 10^{-7}\text{ rad}$.
Q15 — Resolving Power & Optical Instruments · hard · numerical
A microscope uses light of $500\text{ nm}$ with a numerical aperture of $0.5$. Its limit of resolution is:
A. $10^{-6}\text{ m}$
B. $5 \times 10^{-7}\text{ m}$ ✓ Correct
C. $2.5 \times 10^{-7}\text{ m}$
D. $2.5 \times 10^{-6}\text{ m}$
Solution: Limit $= \dfrac{\lambda}{2\,\text{NA}} = \dfrac{5 \times 10^{-7}}{2 \times 0.5} = 5 \times 10^{-7}\text{ m}$.
Q16 — Resolving Power & Optical Instruments · hard · numerical
Two stars separated by $10^{-6}\text{ rad}$ are just resolved using light of $500\text{ nm}$. The diameter of the objective is approximately:
A. $0.61\text{ m}$ ✓ Correct
B. $0.5\text{ m}$
C. $1.22\text{ m}$
D. $2.44\text{ m}$
Solution: $D = \dfrac{1.22\lambda}{\Delta\theta} = \dfrac{1.22 \times 5 \times 10^{-7}}{10^{-6}} = 0.61\text{ m}$.
Q17 — Resolving Power & Optical Instruments · medium · numerical
The wavelength used in a telescope is changed from $400\text{ nm}$ to $600\text{ nm}$. Its limit of resolution becomes:
A. $1.5$ times as small
B. $2.25$ times as large
C. $1.5$ times as large ✓ Correct
D. Unchanged
Solution: Limit of resolution $\propto \lambda$, so it grows in the ratio $\dfrac{600}{400} = 1.5$ — the resolution gets worse.
Q18 — Resolving Power & Optical Instruments · medium · numerical
A telescope aperture of $10\text{ cm}$ is used with light of $500\text{ nm}$. Its limit of resolution is:
A. $6.1 \times 10^{-6}\text{ rad}$ ✓ Correct
B. $1.22 \times 10^{-6}\text{ rad}$
C. $5 \times 10^{-6}\text{ rad}$
D. $6.1 \times 10^{-7}\text{ rad}$
Solution: $\Delta\theta = \dfrac{1.22 \times 5 \times 10^{-7}}{0.1} = 6.1 \times 10^{-6}\text{ rad}$.
Q19 — Resolving Power & Optical Instruments · hard · numerical
A telescope of objective diameter $1.22\text{ m}$ is used with light of $610\text{ nm}$. Its resolving power is approximately:
A. $1.22 \times 10^6$
B. $3.28 \times 10^6$
C. $1.64 \times 10^6$ ✓ Correct
D. $6.1 \times 10^5$
Solution: Resolving power $= \dfrac{D}{1.22\lambda} = \dfrac{1.22}{1.22 \times 6.1 \times 10^{-7}} \approx 1.64 \times 10^6$.
Q20 — Resolving Power & Optical Instruments · hard · numerical
The pupil of the human eye has a diameter of about $2\text{ mm}$. For light of $500\text{ nm}$, the limit of resolution of the eye is approximately:
A. $3.05 \times 10^{-3}\text{ rad}$
B. $3.05 \times 10^{-4}\text{ rad}$ ✓ Correct
C. $6.1 \times 10^{-4}\text{ rad}$
D. $2.5 \times 10^{-4}\text{ rad}$
Solution: $\Delta\theta = \dfrac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}} \approx 3.05 \times 10^{-4}\text{ rad}$.