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JEE Main 2026 (8 Apr, Shift 2) · Resolving Power

Resolving Power question from JEE Main 2026 (8 Apr, Shift 2)

Q4JEE Main 2026 · 8 APR · SHIFT 2Easy

Resolving Power

Some distant star is to be observed by some telescope of diameter of objective lens aa, at an angular resolution of 3.0×10−73.0 \times 10^{-7} radian. If the wavelength of light from the star reaching the telescope is 500 nm, the minimum diameter of the objective lens of the telescope is ______ cm. (nearest interger)

Numerical answer type. Work it out, then open the answer.

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Answer: 203

  1. θ=1.22λa⇒a=1.22λθ\theta = \dfrac{1.22\lambda}{a} \Rightarrow a = \dfrac{1.22\lambda}{\theta}.
  2. a=1.22×500×10−93.0×10−7=2.033a = \dfrac{1.22\times500\times10^{-9}}{3.0\times10^{-7}} = 2.033 m
  3. Minimum diameter ≈203\approx 203 cm.
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