In a Young's double slit experiment,16 fringe | vedclass

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(A) Let the length of the segment be $\ell$.Let $N$ be the number of fringes in the segment $\ell$,and $w$ be the fringe width.The relationship betwee

Vedclass · Accepted Answer

$28$

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(A) Let the length of the segment be $\ell$.Let $N$ be the number of fringes in the segment $\ell$,and $w$ be the fringe width.The relationship between the number of fringes and fringe width is given by $N w = \ell$.Since the fringe width $w = \frac{\lambda D}{d}$,we can write $N \left( \frac{\lambda D}{d} ight) = \ell$.For a fixed segment length $\ell$,the product $N \lambda$ remains constant because $D$ and $d$ are constant.Therefore,$N_1 \lambda_1 = N_2 \lambda_2$.Given $N_1 = 16$,$\lambda_1 = 700 \,nm$,and $\lambda_2 = 400 \,nm$.Substituting the values: $16 	imes 700 = N_2 	imes 400$.$N_2 = \frac{16 	imes 700}{400} = \frac{16 	imes 7}{4} = 4 	imes 7 = 28$.Thus,the number of fringes observed is $28$.

In a Young's double slit experiment,$16$ fringes are observed in a certain segment of the screen when light of wavelength $700 \,nm$ is used. If the wavelength of light is changed to $400 \,nm$,the number of fringes observed in the same segment of the screen would be........

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