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In a biprism experiment, monochromatic light of wavelength ' $\lambda$ ' is used. The distance between two coherent sources ' $\mathrm{d}$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $D_1, D_2, D_3 \& D_4$ and corresponding measured fringe widths are $Z_1, Z_2, Z_3$ and $Z_4$ then
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The correct answer is:
$\frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}$
Fringe width $Z=\frac{\lambda D}{d}$
$\therefore \quad \frac{Z}{D}=\frac{\lambda}{d}=$ constant, as $\lambda$ and $d$ are constant
$$
\therefore \quad \frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}
$$
$\therefore \quad \frac{Z}{D}=\frac{\lambda}{d}=$ constant, as $\lambda$ and $d$ are constant
$$
\therefore \quad \frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}
$$
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