In a Young's double-slit experiment,the fring | vedclass

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

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(A) The formula for fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the screen and the slits,and $d$ is the distance between the slits.Given initial fringe width $\beta_1 = 0.2 \, mm = \frac{\lambda D}{d}$.According to the problem,the new wavelength $\lambda' = \lambda + 0.10\lambda = 1.1\lambda$ and the new slit separation $d' = d + 0.10d = 1.1d$.The new fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda' D}{d'} = \frac{1.1\lambda D}{1.1d} = \frac{\lambda D}{d}$.Since $\frac{\lambda D}{d} = \beta_1$,we have $\beta_2 = \beta_1 = 0.2 \, mm$.

In a Young's double-slit experiment,the fringe width is $0.2 \, mm$. If the wavelength of light is increased by $10\%$ and the distance between the slits is increased by $10\%$,then the new fringe width will be ..... $mm$.

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