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(A) The position of the $n$-th order bright fringe in a Young's double-slit experiment is given by $y_n = n \lambda \frac{D}{d}$.For the third-order b

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$\frac{\lambda_1}{\lambda_2} = \frac{4}{3}$

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(A) The position of the $n$-th order bright fringe in a Young's double-slit experiment is given by $y_n = n \lambda \frac{D}{d}$.For the third-order bright fringe of wavelength $\lambda_1$,the position is $y_1 = 3 \lambda_1 \frac{D}{d}$.For the fourth-order bright fringe of wavelength $\lambda_2$,the position is $y_2 = 4 \lambda_2 \frac{D}{d}$.Given that the fringes coincide,we have $y_1 = y_2$.Therefore,$3 \lambda_1 \frac{D}{d} = 4 \lambda_2 \frac{D}{d}$.Simplifying the equation,we get $3 \lambda_1 = 4 \lambda_2$.Thus,the ratio is $\frac{\lambda_1}{\lambda_2} = \frac{4}{3}$.

Two wavelengths of light $\lambda_1$ and $\lambda_2$ are sent through a Young's double-slit experiment simultaneously. If the third-order bright fringe of $\lambda_1$ coincides with the fourth-order bright fringe of $\lambda_2$,then

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