In a Young's double slit experiment,slits are | vedclass

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(B) For bright fringes to coincide,the path difference must be an integer multiple of both wavelengths. Let $n_1$ be the order for $\lambda_1 = 650 \

Vedclass · Accepted Answer

$7.8$

Vedclass · Answer

(B) For bright fringes to coincide,the path difference must be an integer multiple of both wavelengths. Let $n_1$ be the order for $\lambda_1 = 650 \ nm$ and $n_2$ be the order for $\lambda_2 = 520 \ nm$.The condition for coincidence is $y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$,which implies $n_1 \lambda_1 = n_2 \lambda_2$.$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520 \ nm}{650 \ nm} = \frac{4}{5}$.For the least distance,we take the smallest integers $n_1 = 4$ and $n_2 = 5$.Now,calculate the position $y$ using $n_1 = 4$,$\lambda_1 = 650 	imes 10^{-9} \ m$,$D = 1.5 \ m$,and $d = 0.5 	imes 10^{-3} \ m$:$y = \frac{n_1 \lambda_1 D}{d} = \frac{4 	imes 650 	imes 10^{-9} 	imes 1.5}{0.5 	imes 10^{-3}} \ m$.$y = \frac{4 	imes 650 	imes 1.5}{0.5} 	imes 10^{-6} \ m = 7800 	imes 10^{-6} \ m = 7.8 	imes 10^{-3} \ m = 7.8 \ mm$.

Similar Questions

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The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

Write the formula for fringe width.

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