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A plane wave of wavelength `6250 Å` is incident normally of the principal maximum on a screen distant 50 cm will beA. `312.5 xx 10^(-2)cm`B. `312.5xx10^(-4)cm`C. `312cm`D. `312.5xx10^(-5)cm` |
Answer» Correct Answer - A Here, `lambda = 6250 Å = 6520 x 10^(-10)m` `a = 2xx10^(-2)cm = 2xx10^(-4)m` `D = 50 cm = 0.5 m` `:.` Width of central maxima `= (2lambda D)/(a)` `= (2xx6250xx10^(-10)xx0.5)/(2xx10^(-4)) = 312.5xx10^(-3) cm` |
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