Question 326: Waves Optics

Question

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 '$$\mathrm{D}$$' 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

Accepted Answer

$$\begin{array}{ll} 
& 	ext { Fringe width } Z=\frac{\lambda D}{d} \
	herefore \quad & \frac{Z}{D}=\frac{\lambda}{d}=	ext { constant, as } \lambda 	ext { and } d 	ext { are constant } \
	herefore \quad & \frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}
\end{array}$$

Answer

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

Answer

$$\frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}$$

Answer

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

Answer

$$Z_1 \sqrt{D_1}=Z_2 \sqrt{D_2}=Z_3 \sqrt{D_3}=Z_4 \sqrt{D_4}$$

Waves Optics

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