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Question

$(A)$ The width of the principal maxima in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{a}$, where $\lambda$ is the wavelength

Vedclass · Accepted Answer

$6250$

Vedclass · Answer

$(A)$ The width of the principal maxima in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{a}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $a$ is the slit width.Given: $w = 2.5 	imes 10^{-3}\,m$, $D = 2\,m$, and $a = 1 	imes 10^{-3}\,m$.Substituting the values into the formula: $2.5 	imes 10^{-3} = \frac{2 	imes \lambda 	imes 2}{1 	imes 10^{-3}}$.Solving for $\lambda$: $\lambda = \frac{2.5 	imes 10^{-3} 	imes 10^{-3}}{4} = 0.625 	imes 10^{-6}\,m$.Converting to $\mathring{A}$s: $\lambda = 6.25 	imes 10^{-7}\,m = 6250 	imes 10^{-10}\,m = 6250\,\mathring{A}$.

$A$ parallel beam of monochromatic light is incident on a narrow rectangular slit of width $1\,mm$. When the diffraction pattern is seen on a screen placed at a distance of $2\,m$, the width of the principal maxima is found to be $2.5\,mm$. The wavelength of light is $.............\mathring{A}$.

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