In an interference experiment,the third brigh | vedclass

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(D) The condition for the $n$-th bright fringe in a Young's double-slit experiment is given by $y = \frac{n \lambda D}{d}$.Since the position $y$ on t

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$420$

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(D) The condition for the $n$-th bright fringe in a Young's double-slit experiment is given by $y = \frac{n \lambda D}{d}$.Since the position $y$ on the screen is the same for both cases,we have $y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 3$,$\lambda_1 = 700 \, nm$,and $n_2 = 5$,we substitute these values into the equation:$3 	imes 700 = 5 	imes \lambda_2$.$\lambda_2 = \frac{2100}{5} = 420 \, nm$.Therefore,the required wavelength is $420 \, nm$.

In an interference experiment,the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5$th bright fringe at the same point (in $, nm$)?

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