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Wave Optics question

2019 · 12 Apr · Shift 1 · Q61
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Wave Optics question

2019 · 12 Apr · Shift 1 · Q61

JEE MainPhysicsWave OpticsMCQ+4 / −1
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 Ao\mathop A\limits^oAo​ is used, the minimum separation between two points, to be seen as distinct, will be :
  1. A
    0.12 μ\muμ m
  2. B
    0.38 μ\muμ m
  3. C
    0.24 μ\muμ m
  4. D
    0.48 μ\muμ m
View written solutionFree

Correct answer: C

  1. For a microscope, the minimum resolvable separation according to Rayleigh criterion is

d=0.61λNAd = \frac{0.61\lambda}{\text{NA}}d=NA0.61λ​

where:

  • λ=5000 A˚\lambda = 5000\,\mathring{A}λ=5000A˚
  • NA=1.25\text{NA} = 1.25NA=1.25
  1. Convert wavelength into micrometre:

1 A˚=10−10 m1\,\mathring{A} = 10^{-10}\,\text{m}1A˚=10−10m

So,

5000 A˚=5000×10−10 m=5×10−7 m=0.5 μm5000\,\mathring{A} = 5000 \times 10^{-10}\,\text{m} = 5 \times 10^{-7}\,\text{m} = 0.5\,\mu\text{m}5000A˚=5000×10−10m=5×10−7m=0.5μm

  1. Substitute into the formula:

d=0.61×0.51.25 μmd = \frac{0.61 \times 0.5}{1.25}\,\mu\text{m}d=1.250.61×0.5​μm

d=0.3051.25 μm=0.244 μmd = \frac{0.305}{1.25}\,\mu\text{m} = 0.244\,\mu\text{m}d=1.250.305​μm=0.244μm

  1. Hence,

d≈0.24 μmd \approx 0.24\,\mu\text{m}d≈0.24μm

  1. Checking options:
  • A: 0.12 μ0.12\,\mu0.12μm
  • B: 0.38 μ0.38\,\mu0.38μm
  • C: 0.24 μ0.24\,\mu0.24μm
  • D: 0.48 μ0.48\,\mu0.48μm

So the correct option is C.

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