In a Young's double slit experiment,12 fringe | vedclass

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(B) The width of the segment on the screen is given by $x = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Since the segment length

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$18$

Vedclass · Answer

(B) The width of the segment on the screen is given by $x = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Since the segment length $x$ remains constant,we have $n_1 \beta_1 = n_2 \beta_2$.Substituting the formula for fringe width: $n_1 \left( \frac{\lambda_1 D}{d} ight) = n_2 \left( \frac{\lambda_2 D}{d} ight)$.This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 12$,$\lambda_1 = 600 \, nm$,and $\lambda_2 = 400 \, nm$,we find $n_2 = \frac{n_1 \lambda_1}{\lambda_2}$.$n_2 = \frac{12 	imes 600}{400} = 12 	imes 1.5 = 18$.Therefore,the number of fringes observed is $18$.

Column-$I$	Column-$II$
$(a)$ Gymnosperms	$(p)$ Trimerous flowers
$(b)$ Monocots	$(q)$ Absence of flowers
$(c)$ Dicots	$(r)$ Primitive flowers
$(d)$ Pteridophytes	$(s)$ Tetramerous or pentamerous flowers

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen,when light of wavelength $600 \, nm$ is used. If the wavelength of light is changed to $400 \, nm$,the number of fringes observed in the same segment of the screen will be

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