Physics - Wave Optics Question with Solution | TestHub

Let$n\text{th}$minimum of 400nm coincides with$m\text{th}$minimum of 560 nm, then

For${\mathrm{n}}^{\mathrm{th}}$minima$\mathrm{\delta }=\left(2\mathrm{n}-1\right)\frac{\mathrm{\lambda }}{2}$

$\left(2n-1\right)\left(\frac{400}{2}\right)=\left(2\text{m}-1\right)\left(\frac{560}{2}\right)$

$\frac{2n-1}{2\text{m}-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}$

i.e.,$4\text{th}$minimum of 400 nm coindes with 3rd minimum of 560 nm.

Distance of${\mathrm{n}}^{\mathrm{th}}$minima from center$=\frac{\left(2\mathrm{n}-1\right)\mathrm{\lambda D}}{2\mathrm{d}}$

Location of this minimum is,

${\text{Y}}_1=\frac{\left(2\times 4-1\right)\left(1000\right)\left(400\times 10^{-6}\right)}{2\times 0·1}=14\text{mm}$

Next 11th minimum of 400 nm will coincide with 8th
minimum of 560 nm.

Location of this minimum is,

${\text{Y}}_2=\frac{\left(2\times 11-1\right)\left(1000\right)\left(400\times 10^{-6}\right)}{2\times 0·1}=42\text{mm}$

$\therefore$Required distance${=\text{Y}}_2-{\text{Y}}_1=28\text{mm}$

Hence, the correct option is (28 mm)