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JEE Main 2016 (10 Apr, Shift 1) · Resolving Power

Resolving Power question from JEE Main 2016 (10 Apr, Shift 1)

Q21JEE Main 2016 · 10 APR · SHIFT 1Medium

Resolving Power

Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year = 9.46 ×\times 1015 m) of the order of:

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Answer: (b)

  1. θ=1.22λD=1.22×600×10−90.3=2.44×10−6\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times600\times10^{-9}}{0.3} = 2.44\times10^{-6} rad
  2. L=10×9.46×1015=9.46×1016L = 10 \times 9.46\times10^{15} = 9.46\times10^{16} m
  3. x=Lθ=9.46×1016×2.44×10−6≈2.3×1011x = L\theta = 9.46\times10^{16}\times2.44\times10^{-6} \approx 2.3\times10^{11} m =2.3×108= 2.3\times10^{8} km
  4. So the answer is (b) of the order of 10810^{8} km.
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