${y}_{1}={A}_{1} \sin {k}({x}-v {t}), {y}_{2}={A}_{2} \sin {k}\left({x}-{vt}+{x}_{0}\right) .$ Given amplitudes ${A}_{1}=12\, {mm}$ and ${A}_{2}=5\, {mm}$ ${x}_{0}=3.5\, {cm}$ and wave number ${k}=6.28\, {cm}^{-1}$. The amplitude of resulting wave will be $......\,{mm}$
${y}_{1}=12 \sin 6.28({x}-{vt})$
${y}_{2}=5 \sin 6.28({x}-{vt}+3.5)$
$\Delta \phi=\frac{2 \pi}{\lambda}(\Delta {x})$
$={K}(\Delta {x})$
$=6.28 \times 3.5=\frac{7}{2} \times 2 \pi=7 \pi$
${A}_{{DE}}=\sqrt{{A}_{1}^{2}+{A}_{2}^{2}+2 {A}_{1} {A}_{2} \cos \phi}$
${A}_{{DA}}=\sqrt{(12)^{2}+(5)^{2}+2(12)(5) \cos (7 \pi)}$
$=\sqrt{144+25-120}$
$=\,7$
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light. If distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is.....$mm$
