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Question #43591

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The Question

A telescope with a circular objective lens of diameter $D = 0.122 \text{ m}$ is used to observe two distant point sources emitting light of frequency $f_0 = 6 \times 10^{14} \text{ Hz}$ in a vacuum. The entire setup (sources and telescope) is then completely immersed in a transparent, non-magnetic liquid of dielectric constant $K = 4$ . Let $\Delta \theta$ be the new minimum angular separation required to just resolve the two sources according to the Rayleigh criterion, and $f$ be the frequency of the light in the liquid. Which of the following represents the correct values for $\Delta \theta$ and $f$ ? (Take the speed of light in vacuum as $c = 3 \times 10^8 \text{ m/s}$ )

Options

A.
$\Delta \theta = 2.5 \times 10^{-6} \text{ rad}, \; f = 6 \times 10^{14} \text{ Hz}$
B.
$\Delta \theta = 5.0 \times 10^{-6} \text{ rad}, \; f = 3 \times 10^{14} \text{ Hz}$
C.
$\Delta \theta = 1.0 \times 10^{-5} \text{ rad}, \; f = 6 \times 10^{14} \text{ Hz}$
D.
$\Delta \theta = 2.5 \times 10^{-6} \text{ rad}, \; f = 3 \times 10^{14} \text{ Hz}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Angular Resolution Limit|Spectrum Order and Frequency/Wavelength|Dielectric Constant (K) Shift

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of wave properties and the Rayleigh criterion in different optical media.

A common trap is assuming that the frequency of light changes when it enters a dielectric medium, leading students to incorrectly calculate the wavelength or select an option with a modified frequency.

Another trap is forgetting that the angular resolution limit ( $\Delta \theta \approx 1.22\lambda/D$ ) depends strictly on the wavelength in the medium ( $\lambda_m = \lambda_0/\sqrt{K}$ ), not the vacuum wavelength.

Students who misapply the relation $v = f\lambda$ by keeping $\lambda$ constant and changing $f$ , or who multiply the wavelength by the refractive index instead of dividing, will fall for the distractor options.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Angular Resolution Limit
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free