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Wave Optics question

2015 · Shift 0 · Q48
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Wave Optics question

2015 · Shift 0 · Q48

JEE MainPhysicsWave OpticsMCQ+4 / −1
Assuming human pupil to have a radius of 0.25cm0.25cm0.25cm and a comfortable viewing distance of 25cm25cm25cm, the minimum separation between two objects that human eye can resolve at 500nm500nm500nm wavelength is :
  1. A
    100 μm100\,\mu m100μm
  2. B
    300 μm300\,\mu m300μm
  3. C
    1 μm1\,\mu m1μm
  4. D
    30 μm30\,\mu m30μm
View written solutionFree

Correct answer: D

  1. Use Rayleigh criterion for a circular aperture

For a circular aperture (human pupil), the minimum angular resolution is

θmin⁡=1.22λD\theta_{\min} = 1.22\frac{\lambda}{D}θmin​=1.22Dλ​

where:

  • λ=500 nm=5×10−7 m\lambda = 500\,\text{nm} = 5\times 10^{-7}\,\text{m}λ=500nm=5×10−7m
  • pupil radius r=0.25 cm=2.5×10−3 mr = 0.25\,\text{cm} = 2.5\times 10^{-3}\,\text{m}r=0.25cm=2.5×10−3m
  • so diameter D=2r=0.5 cm=5×10−3 mD = 2r = 0.5\,\text{cm} = 5\times 10^{-3}\,\text{m}D=2r=0.5cm=5×10−3m
  1. Calculate angular resolution
θmin⁡=1.225×10−75×10−3=1.22×10−4 rad\theta_{\min} = 1.22\frac{5\times 10^{-7}}{5\times 10^{-3}} = 1.22\times 10^{-4}\,\text{rad}θmin​=1.225×10−35×10−7​=1.22×10−4rad
  1. Convert angular resolution to linear separation at the near point

Comfortable viewing distance:

L=25 cm=0.25 mL = 25\,\text{cm} = 0.25\,\text{m}L=25cm=0.25m

For small angles, minimum resolvable separation is

s=Lθmin⁡s = L\theta_{\min}s=Lθmin​

So,

s=0.25×1.22×10−4=3.05×10−5 ms = 0.25\times 1.22\times 10^{-4} = 3.05\times 10^{-5}\,\text{m}s=0.25×1.22×10−4=3.05×10−5m
  1. Convert into micrometres
3.05×10−5 m=30.5×10−6 m=30.5 μm3.05\times 10^{-5}\,\text{m} = 30.5\times 10^{-6}\,\text{m} = 30.5\,\mu m3.05×10−5m=30.5×10−6m=30.5μm

Thus the minimum separation is approximately

30 μm\boxed{30\,\mu m}30μm​
  1. Check options
  • A: 100 μm100\,\mu m100μm ❌
  • B: 300 μm300\,\mu m300μm ❌
  • C: 1 μm1\,\mu m1μm ❌
  • D: 30 μm30\,\mu m30μm ✅

Therefore, the correct option is D.

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