In a special arrangement of Young's double-slit experiment, the...

From diagram as provided in question, $OP = x$
$CO = D$
${S}_{1}C = {S}_{2}C = D{T}_{1}P = {T}_{1}O - OP = D - x$
${T}_{2}P = {T}_{2}O + OP = D + x$
Now, ${S}_{1}p = \sqrt{{\left({S}_{1}{T}_{1}\right)}^{2} + {\left({T}_{1}P\right)}^{2}} = \sqrt{{D}^{2} + {\left(D - x\right)}^{2}}$

${S}_{2}P = \sqrt{{\left({S}_{2}{T}_{2}\right)}^{2} + {\left({T}_{2}P\right)}^{2}} = \sqrt{{D}^{2} + {\left(D + x\right)}^{2}}$

For fist minimum to occur,

Path difference

${S}_{2}P - {S}_{1}P = \dfrac{\lambda}{2}$

$⇒ \sqrt{{D}^{2} + {\left(D + x\right)}^{2}} - \sqrt{{D}^{2} + {\left(D - x\right)}^{2}} = \dfrac{\lambda}{2}$

The first minimum falls at a distance $D$ from the centre, i.e., $x = D$

\left[{D}^{2} + 4{D}^{2}\right]\dfrac{1}{2} - D = \dfrac{\lambda}{2}$</p><p>$∴ D\left(\sqrt{5} - 1\right) = \dfrac{\lambda}{2}$</p><p>$⇒ D = \dfrac{\lambda}{2\left(2 . 24 - 1\right)} = \dfrac{\lambda}{2 . 48}