Physics - Wave Optics Question with Solution | TestHub

Given${\text{Y}}_1=40{\text{ cm, D}}_1=2\text{m}=200\text{cm},D_2=10\text{cm}$

$\text{tan}\alpha =\frac{y_1}{{\text{D}}_1}=\frac{40}{200}=\frac{1}{5}$

$\therefore \alpha ={\text{tan}}^{-1}\left(1/5\right)$

$\text{sin}\alpha =\frac{1}{\sqrt{26}}\approx \frac{1}{5}=\text{tan}\alpha$



Initial path difference${\text{Δx}}_{\text{1}},={\text{ SS}}_1-{\text{SS}}_2=\frac{{\text{y}}_{\text{1}}\text{d}}{{\text{D}}_{\text{1}}}$=$\frac{{\text{y}}_1\text{d}}{{\text{D}}_1}$& geometrical path difference$\Delta {\text{x}}_2$${\text{S}}_{\text{1}}\text{R}-{\text{S}}_{\text{2}}\text{R}=$$-\frac{{\text{y}}_2\text{d}}{{\text{D}}_2}$

For central maxima, total path difference has to be zero.

i.e.,$\frac{{\text{y}}_1\text{d}}{{\text{D}}_1}-\frac{{\text{y}}_2\text{d}}{{\text{D}}_2}=0$

$\Rightarrow {\text{y}}_2={\text{y}}_1\frac{{\text{D}}_2}{{\text{D}}_1}=40\text{cm}\times \frac{10\text{cm}}{200\text{cm}}=2\text{cm}$

Now as central bright fringe is at Q. them the optical path difference is$\left(\mu -1\right)\text{t}$& for central maxima$\left(\mu -1\right)\text{t}=\Delta {\text{x}}_1$

$\mu =1+\frac{\Delta {\text{x}}_1}{\text{t}}=1+\frac{{\text{y}}_1\text{d}}{{\text{t D}}_1}$

$=1+\frac{0\text{.}4\text{m}\times 0\text{.}8\text{mm}}{0\text{.}1\text{m}\times 2\text{m}}$

$=1+2\times 0\text{.}8\times 10^{-3}$

$=1+1\text{.}6\times 10^{-3}$

$=1\text{.}0016$