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Wave Optics question

2019 · 9 Apr · Shift 2 · Q61
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Wave Optics question

2019 · 9 Apr · Shift 2 · Q61

JEE MainPhysicsWave OpticsMCQ+4 / −1
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600nm. coming from a distant object, the limit of resolution of the telescope is close to :-
  1. A
    3.0 × 10–7 rad
  2. B
    4.5 × 10–7 rad
  3. C
    1.5 × 10–7 rad
  4. D
    2.0 × 10–7 rad
View written solutionFree

Correct answer: A

  1. Use Rayleigh criterion for a circular aperture

For a telescope, the angular limit of resolution is

θ=1.22λD\theta = 1.22\frac{\lambda}{D}θ=1.22Dλ​

where:

  • λ=600 nm=600×10−9 m\lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m}λ=600nm=600×10−9m
  • D=250 cm=2.5 mD = 250\,\text{cm} = 2.5\,\text{m}D=250cm=2.5m
  1. Substitute the values

θ=1.22600×10−92.5\theta = 1.22\frac{600 \times 10^{-9}}{2.5}θ=1.222.5600×10−9​

θ=1.22×240×10−9\theta = 1.22 \times 240 \times 10^{-9}θ=1.22×240×10−9

θ=292.8×10−9 rad\theta = 292.8 \times 10^{-9} \text{ rad}θ=292.8×10−9 rad

θ=2.928×10−7 rad\theta = 2.928 \times 10^{-7} \text{ rad}θ=2.928×10−7 rad

  1. Approximate

θ≈3.0×10−7 rad\theta \approx 3.0 \times 10^{-7}\,\text{rad}θ≈3.0×10−7rad

  1. Match with options
  • A: 3.0×10−7 rad3.0 \times 10^{-7}\,\text{rad}3.0×10−7rad ✅
  • B: 4.5×10−7 rad4.5 \times 10^{-7}\,\text{rad}4.5×10−7rad
  • C: 1.5×10−7 rad1.5 \times 10^{-7}\,\text{rad}1.5×10−7rad
  • D: 2.0×10−7 rad2.0 \times 10^{-7}\,\text{rad}2.0×10−7rad

Therefore, the correct option is A.

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