In a biprism experiment, the slit separation is $1 \mathrm{~mm}$. Using monochromatic light of wavelength $5000 Å$, an interference pattern is obtained on the screen. Where should the screen be moved, so that the change in fringe width is $12 \cdot 5 imes 10^{-5} \mathrm{~m}$ ?

Question

In a biprism experiment, the slit separation is $1 \mathrm{~mm}$. Using monochromatic light of wavelength $5000 Å$, an interference pattern is obtained on the screen. Where should the screen be moved, so that the change in fringe width is $12 \cdot 5 	imes 10^{-5} \mathrm{~m}$ ?

Aicharya · Accepted Answer

$\mathrm{d}=1 	imes 10^{-3} \mathrm{~m}$ $\lambda=5000 Å=5 	imes 10^{-7} \mathrm{~m}$ $\beta_{2}-\beta_{1}=12.5 	imes 10^{-5} \mathrm{~m}$ $\mathrm{D}_{2}-\mathrm{D}_{1}=?$ $\beta_{1}=\frac{\lambda \mathrm{D}_{1}}{\mathrm{~d}} \quad \beta_{2}=\frac{\lambda \mathrm{D}_{2}}{\mathrm{~d}}$ $\beta_{2}-\beta_{1}=\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_{2}-\mathrm{D}_{1}ight)$ $	herefore \quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}ight)=\frac{\mathrm{d}}{\lambda}\left(\beta_{2}-\beta_{1}ight)$ $\quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}ight)=\frac{10^{-3}}{5 	imes 10^{-7}} 	imes 12.5 	imes 10^{-5}=\frac{12.5}{50}=25 \mathrm{~cm}$

In a biprism experiment, the slit separation is $1 \mathrm{~mm}$. Using monochromatic light of wavelength…

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