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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 $12 \cdot 5 \times 10^{-5} \mathrm{~m}$ ?
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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 $12 \cdot 5 \times 10^{-5} \mathrm{~m}$ ?
Solution:
1792 Upvotes
Verified Answer
The correct answer is:
Away or towards the slit by 25 am
$\mathrm{d}=1 \times 10^{-3} \mathrm{~m}$
$\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}$
$\beta_{2}-\beta_{1}=12.5 \times 10^{-5} \mathrm{~m}$
$\mathrm{D}_{2}-\mathrm{D}_{1}=?$
$\beta_{1}=\frac{\lambda \mathrm{D}_{1}}{\mathrm{~d}} \quad \beta_{2}=\frac{\lambda \mathrm{D}_{2}}{\mathrm{~d}}$
$\beta_{2}-\beta_{1}=\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)$
$\therefore \quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{\mathrm{d}}{\lambda}\left(\beta_{2}-\beta_{1}\right)$
$\quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{10^{-3}}{5 \times 10^{-7}} \times 12.5 \times 10^{-5}=\frac{12.5}{50}=25 \mathrm{~cm}$
$\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}$
$\beta_{2}-\beta_{1}=12.5 \times 10^{-5} \mathrm{~m}$
$\mathrm{D}_{2}-\mathrm{D}_{1}=?$
$\beta_{1}=\frac{\lambda \mathrm{D}_{1}}{\mathrm{~d}} \quad \beta_{2}=\frac{\lambda \mathrm{D}_{2}}{\mathrm{~d}}$
$\beta_{2}-\beta_{1}=\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)$
$\therefore \quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{\mathrm{d}}{\lambda}\left(\beta_{2}-\beta_{1}\right)$
$\quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{10^{-3}}{5 \times 10^{-7}} \times 12.5 \times 10^{-5}=\frac{12.5}{50}=25 \mathrm{~cm}$
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