A screen is placed 50 , cm from a single slit | vedclass

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(B) The condition for the $n^{th}$ minima in a single-slit diffraction pattern is given by $y_n = n \lambda \frac{D}{a}$, where $D$ is the distance to

Vedclass · Accepted Answer

$0.2$

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(B) The condition for the $n^{th}$ minima in a single-slit diffraction pattern is given by $y_n = n \lambda \frac{D}{a}$, where $D$ is the distance to the screen, $a$ is the slit width, and $\lambda$ is the wavelength.Given: $D = 50 \, cm = 0.5 \, m$, $\lambda = 6000 \, \mathring{A} = 6 	imes 10^{-7} \, m$, and the distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta y = 3 \, mm = 3 	imes 10^{-3} \, m$.The distance between the $n_1^{th}$ and $n_2^{th}$ minima is $\Delta y = (n_2 - n_1) \lambda \frac{D}{a}$.Substituting the values: $3 	imes 10^{-3} = (3 - 1) 	imes (6 	imes 10^{-7}) 	imes \frac{0.5}{a}$.$3 	imes 10^{-3} = 2 	imes 6 	imes 10^{-7} 	imes \frac{0.5}{a} = 6 	imes 10^{-7} 	imes \frac{1}{a}$.$a = \frac{6 	imes 10^{-7}}{3 	imes 10^{-3}} = 2 	imes 10^{-4} \, m = 0.2 \, mm$.

$A$ screen is placed $50 \, cm$ from a single slit, which is illuminated with $6000 \, \mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3 \, mm$, the width of the slit is.....$mm$

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