In Young's double slit experiment, slits are separated by \( 2 \mathrm{~mm} \) and the screen is placed at a
distance of \( 1.2 \mathrm{~m} \) from the slits. Light consisting of two wavelengths \( 6500 Å \) and \( 5200 Å \) are
used to obtain interference fringes. Then the separation between the fourth bright fringes of
two different patterns produced by the two wavelengths is

We know that,
\(y=\frac{n \lambda D}{d}\)
where \(\lambda\) is wavelength; \(\mathrm{D}\) is distance of screen from slits; \(\mathrm{d}\) is separation between the slits. So,
\(y_{1}=\frac{n \lambda_{1} D}{d}\) and \(y_{2}=\frac{n \lambda_{2} D}{d}\)
Therefore, \(y_{1}-y_{2}=\frac{n D}{d}\left(\lambda_{1}-\lambda_{2}\right)\)
Given,
\(n=4 ; D=1.2 \mathrm{~m} ; d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} ; \lambda_{1}=6500 Å=6500 \times 10^{-10} \mathrm{~m} ; \lambda_{2}=5200 Å=5200 \times 10^{-10} \mathrm{~m}\)
\(y_{1}-y_{2}=\frac{4 \times 1.2}{2 \times 10^{-3}}\left(6500 \times 10^{-10}-5200 \times 10^{-10}\right)\)
\(=2.4 \times 10^{3} \times 1300 \times 10^{-10} \mathrm{~m}\)
\(=3120 \times 10^{-7}=0.3120 \times 10^{-3} \mathrm{~m}\)
\(\Rightarrow y_{1}-y_{2}=0.3120 \mathrm{~mm}\)
Therefore, the separation between the fourth bright fringesof two different patterns produced by the two wavelengths is
\(0.3120 \mathrm{~mm} .\)