Physics - Wave Optics Question with Solution | TestHub

The fringe width in a biprism experiment is given by the formula:

$W=\frac{\gamma D}{d}$

where $\gamma$ is the wavelength of monochromatic light, $D$ is the distance between the slit and the eyepiece, and $d$ is the distance between the two coherent sources.

Since $\gamma$ and $d$ are kept constant, the fringe width $W$ is directly proportional to the distance $D$.

$W\propto D$

$\Rightarrow \frac{W}{D}=\text{constant}$

Therefore, for different distances $D_1,D_2,D_3,D_4$ and corresponding fringe widths $W_1,W_2,W_3,W_4$, we have:

$\frac{W_1}{D_1}=\frac{W_2}{D_2}=\frac{W_3}{D_3}=\frac{W_4}{D_4}$

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