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If $a_{1}$ and $a_{2}$ are the amplitudes of two sources in Young's double slit experiment, then the maximum intensity of interference fringe is(a) $left(a_{1}...

Question

If $a_{1}$ and $a_{2}$ are the amplitudes of two sources in Young's double slit experiment, then the maximum intensity of interference fringe is(a) $left(a_{1}+a_{2}ight)$(b) $2left(a_{1}+a_{2}ight)$(c) $left(a_{1}+a_{2}ight)^{2}$(d) $left(a_{1}-a_{2}ight)^{2}$

If $a_{1}$ and $a_{2}$ are the amplitudes of two sources in Young's double slit experiment, then the maximum intensity of interference fringe is (a) $left(a_{1}+a_{2} ight)$ (b) $2left(a_{1}+a_{2} ight)$ (c) $left(a_{1}+a_{2} ight)^{2}$ (d) $left(a_{1}-a_{2} ight)^{2}$



Answers

In the Young's double slit experiment, the two slits are identical (each having intensity $\mathrm{I}_{0}$ ). The intensity of the central maximum is (a) $2 \mathrm{I}_{0}$ (b) $4 \mathrm{I}_{0}$ (c) I (d) $3 \mathrm{I}_{0}$


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