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(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is

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$6000$

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(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the slit separation.When the screen is moved by $\Delta D = 5 	imes 10^{-2} \, m$, the change in fringe width is $\Delta \beta = 3 	imes 10^{-5} \, m$.From the formula, the change in fringe width is $\Delta \beta = \frac{\lambda}{d} \Delta D$.Rearranging for $\lambda$, we get $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.Substituting the given values:$\lambda = \frac{(3 	imes 10^{-5} \, m) 	imes (10^{-3} \, m)}{5 	imes 10^{-2} \, m}$.$\lambda = \frac{3 	imes 10^{-8}}{5 	imes 10^{-2}} \, m = 0.6 	imes 10^{-6} \, m = 6 	imes 10^{-7} \, m$.Converting to $	ext{Å}$ $(1 \, 	ext{Å} = 10^{-10} \, m)$:$\lambda = 6000 	imes 10^{-10} \, m = 6000 \, 	ext{Å}$.

Similar Questions

In which of the following is the interference due to the division of wave front?

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

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