Question
Using Young's double slit experiment, a monochromatic light of wavelength $5000\,\mathring A$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000\,\mathring A$ is used and the separation between the slits is doubled, then the new fringe width will be ............... $mm$

Answer

Fringe width $\beta=\frac{ D \lambda}{ d }$

$\lambda_{1}=5000 \,\mathring A$

$\beta_{1}=\frac{ D }{ d }\left(5000 \times 10^{-10}\right)=5 \times 10^{-4} \,m$......$(I)$

$\beta_{2}=\frac{ D }{(2 d )}\left(6000 \times 10^{-10}\right)= x \text { (let) }$.....$(II)$

Divide $(II)$ and $(I)$

$\frac{\beta_{2}}{\beta_{1}}=\frac{3000 \times 10^{-10}}{5000 \times 10^{-10}}=\frac{x}{5 \times 10^{-4}}$

$x=3 \times 10^{-4} m \text { or } 0.3 \,mm$

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