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(B) The condition for the $n^{th}$ bright fringe of red light to coincide with the $(n+1)^{th}$ bright fringe of blue light is given by equating their

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$2$

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(B) The condition for the $n^{th}$ bright fringe of red light to coincide with the $(n+1)^{th}$ bright fringe of blue light is given by equating their positions on the screen.The position of the $n^{th}$ bright fringe is $y_n = \frac{n \lambda D}{d}$.For red light: $y_{n, red} = \frac{n \lambda_{red} D}{d}$.For blue light: $y_{n+1, blue} = \frac{(n+1) \lambda_{blue} D}{d}$.Equating the two positions: $\frac{n \lambda_{red} D}{d} = \frac{(n+1) \lambda_{blue} D}{d}$.This simplifies to: $n \lambda_{red} = (n+1) \lambda_{blue}$.Substituting the given values: $n(7800) = (n+1)(5200)$.Dividing both sides by $2600$: $3n = 2(n+1)$.$3n = 2n + 2$.$n = 2$.

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

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