A vessel ABCD of 10 cm width has two small slits S _ 1 and S _ 2...

Given ${\text{Y}}_{1} = 40{\text{cm, D}}_{1} = 2\text{m} = 200\text{cm}, {D}_{2} = 10\text{cm}$

$\text{tan} \alpha = \dfrac{{y}_{1}}{{\text{D}}_{1}} = \dfrac{4 0}{2 0 0} = \dfrac{1}{5}$

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

$\text{sin} \alpha = \dfrac{1}{\sqrt{2 6}} \approx \dfrac{1}{5} = \text{tan} \alpha$



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

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

i.e., $\dfrac{{\text{y}}_{1} \text{d}}{{\text{D}}_{1}} - \dfrac{{\text{y}}_{2} \text{d}}{{\text{D}}_{2}} = 0$

$⇒ {\text{y}}_{2} = {\text{y}}_{1} \dfrac{{\text{D}}_{2}}{{\text{D}}_{1}} = 4 0 \text{cm} \times \dfrac{1 0 \text{cm}}{2 0 0 \text{cm}} = 2 \text{cm}$

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}$<br /><br />$\mu = 1 + \dfrac{\Delta {\text{x}}_{1}}{\text{t}} = 1 + \dfrac{{\text{y}}_{1} \text{d}}{{\text{t D}}_{1}}$<br /><br />$= 1 + \dfrac{0 \text{.} 4 \text{m} \times 0 \text{.} 8 \text{mm}}{0 \text{.} 1 \text{m} \times 2 \text{m}}$<br /><br />$= 1 + 2 \times 0 \text{.} 8 \times 1 {0}^{- 3}$<br /><br />$= 1 + 1 \text{.} 6 \times 1 {0}^{- 3}$<br /><br />$= 1 \text{.} 0 0 1 6