Question 304: Waves Optics

Question

In a biprism experiment, monochromatic light of wavelength ' $\gamma$ ' is used. The distance between the two coherent sources ' $d$ ' 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 $\mathrm{W}_1, \mathrm{~W}_2, \mathrm{~W}_3, \mathrm{~W}_4$ then

Accepted Answer

In a biprism experiment using monochromatic light with wavelength ' $\lambda$ ', the distance ' $d$ ' between the two coherent sources remains constant. As you change the distance between the slit and the eyepiece to $D_1, D_2, D_3, D_4$, the corresponding fringe widths are measured as $W_1, W_2, W_3, W_4$.
The formula for the fringe width $\mathrm{W}$ is:
$  \mathrm{W} = \frac{\lambda \mathrm{D}}{d} $
From this equation, we can derive:
$  \frac{\mathrm{W}}{\mathrm{D}} = \frac{\lambda}{d} $
Since $\lambda$ and $d$ are constants, $\frac{\mathrm{W}}{\mathrm{D}}$ is also a constant. Therefore, this relationship holds for all the measured pairs:
$  \frac{W_1}{D_1} = \frac{W_2}{D_2} = \frac{W_3}{D_3} = \frac{W_4}{D_4} $

Answer

$\mathrm{W}_1 \mathrm{D}_1=\mathrm{W}_2 \mathrm{D}_2=\mathrm{W}_3 \mathrm{D}_3=\mathrm{W}_4 \mathrm{D}_4$

Answer

$\frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}$

Answer

$\mathrm{W}_1 \sqrt{\mathrm{D}_1}=\mathrm{W}_2 \sqrt{\mathrm{D}_2}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}$

Answer

$\mathrm{D}_1 \sqrt{\mathrm{~W}_1}=\mathrm{D}_2 \sqrt{\mathrm{~W}_2}=\mathrm{D}_3 \sqrt{\mathrm{~W}_3}=\mathrm{D}_4 \sqrt{\mathrm{~W}_4}$

Waves Optics

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