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Multiple choice questions · Q12

Q.Two point white dots are 1 mm apart on a black paper. They are viewed by an eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ=500\lambda = 500 nm]

(a) 1 m
(b) 5 m
(c) 3 m
(d) 6 m
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Step 1. Given: dot separation s=1s=1 mm =1×10−3=1\times10^{-3} m, pupil diameter a=3a=3 mm =3×10−3=3\times10^{-3} m, λ=500\lambda=500 nm =5×10−7=5\times10^{-7} m.

Step 2. Rayleigh's criterion gives the minimum resolvable angular separation as θ=1.22λ/a=1.22×5×10−7/(3×10−3)=2.03×10−4\theta=1.22\lambda/a=1.22\times5\times10^{-7}/(3\times10^{-3})=2.03\times10^{-4} rad.

Step 3. For small angles, θ≈s/D\theta\approx s/D, where DD is the (maximum) distance at which the two dots are still just resolvable; rearranging, D=s/θ=(1×10−3)/(2.03×10−4)≈4.9D=s/\theta=(1\times10^{-3})/(2.03\times10^{-4})\approx4.9 m, which rounds to the given answer of 5 m. …

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