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(D) Let the order of the bright fringe for the longer wavelength $\lambda_1 = 650 \, nm$ be $n_1$ and for the shorter wavelength $\lambda_2 = 520 \, n

Vedclass · Accepted Answer

$4$

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(D) Let the order of the bright fringe for the longer wavelength $\lambda_1 = 650 \, nm$ be $n_1$ and for the shorter wavelength $\lambda_2 = 520 \, nm$ be $n_2$.At the position of coincidence,the distance from the central maximum is the same for both fringes:$y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$n_1 \lambda_1 = n_2 \lambda_2$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650}$Simplifying the ratio: $\frac{n_1}{n_2} = \frac{52}{65} = \frac{4}{5}$Thus,the order of the bright fringe for the longer wavelength is $n_1 = 4$ and for the shorter wavelength is $n_2 = 5$.The question asks for the order of the bright fringe of the longer wavelength,which is $4$.

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