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

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

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(A) The formula for fringe width in a 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 separation between the slits.Given that the initial fringe width $\beta_1 = 0.2 \, mm$.The new wavelength $\lambda' = \lambda + 0.10\lambda = 1.10\lambda$.The new slit separation $d' = d + 0.10d = 1.10d$.The new fringe width $\beta_2 = \frac{\lambda' D}{d'} = \frac{(1.10\lambda) D}{(1.10d)} = \frac{\lambda D}{d} = \beta_1$.Therefore,the fringe width remains unchanged at $0.2 \, mm$.

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

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