An electron with kinetic energy K₁ enters between parallel plates of a capacitor at an angle $' α'$ with the plates. It leaves the plates at angle $' β'$ with kinetic energy K₂ . Then the ratio of kinetic energies $K_{1} : K_{2}$ will be :

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$\frac{\cos^{2} β}{\cos^{2} α}$

$\frac{\cos β}{\sin α}$

$\frac{\cos β}{\cos α}$

$\frac{\sin^{2} β}{\cos^{2} α}$

answer is C.

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Detailed Solution

From figure, $v_{1} \cos α = v_{2} \cos β$
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$∴ \frac{K_{1}}{K_{2}} = \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{\cos^{2} β}{\cos^{2} α}$

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