Question

Rayleigh’s criterion is used to determine when two objects are barely resolved by a lens of diameter d. The angular separation must be greater than θR\theta_{R} where θR=1.22λ/d\theta_{R} =1.22 \lambda /d . In order to resolve two objects 4000 nm apart at a distance of 20 cm with a lens of diameter 5 cm, what energy (a) photons or (b) electrons should be used? Is this consistent with the uncertainty principle?

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Given:\textbf{Given:}

Diameter of the lens (d) = 55 cm

The distance at which the two objects is resolved (xx) = 2020 cm

The resulting distance in separation (yy) = 40004000 nm

θR=1.22λd\theta_R=1.22\dfrac{\lambda}{d}


Since the minimum angle, at which the two objects are resolved, is given as follows:

tanθR=yx          and          θRtanθR            for small     θ\begin{equation*} \tan \theta _R =\dfrac{y}{x} \;\;\;\;\;\text{and}\;\;\;\;\;\theta _R \simeq \tan \theta _R \;\;\;\;\;\;\textbf{for small\;\; $\theta$} \end{equation*}

therefore,

θR=yx=4000×109  m20×102  m=2×105\begin{align*} \theta_R&=\dfrac{y}{x}\\\\ &=\dfrac{4000\times {10}^{-9}\;\cancel{\text{m}}}{20\times {10}^{-2}\;\cancel{m}}\\\\ &=2\times {10}^{-5} \end{align*}

and since

θR=1.22λd\begin{align*} \theta_R=1.22\dfrac{\lambda}{d} \end{align*}

therefore, the minimum wavelength for a given particle is given as follows:

λmin=dθR1.22=5×102  m×2×1051.228.2×107  m\begin{align*} \lambda_{\text{min}} &=\dfrac{d\cdot \theta_R}{1.22}\\\\ &=\dfrac{5 \times {10}^{-2}\;\text{m}\times 2 \times {10}^{-5}}{1.22}\\\\ &\simeq 8.2 \times {10}^{-7}\;\text{m} \end{align*}

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