In Young's double-slit experiment,the fringe | 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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$0.2$

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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 slits and the screen,and $d$ is the distance between the two slits.Given that the wavelength $\lambda$ increases by $10\%$,the new wavelength $\lambda' = \lambda + 0.1\lambda = 1.1\lambda$.The distance between the slits $d$ increases by $10\%$,so the new distance $d' = d + 0.1d = 1.1d$.The new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d'} = \frac{(1.1\lambda) D}{(1.1d)} = \frac{\lambda D}{d} = \beta$.Since $\beta = 0.2 \, mm$,the new fringe width $\beta'$ remains $0.2 \, mm$.

In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

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