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(D) For a single slit diffraction, the condition for the $n^{	ext{th}}$ secondary maxima is given by the formula:
$a \sin 	heta = (2n + 1) \frac{\la

Vedclass · Accepted Answer

$2748$

Vedclass · Answer

(D) For a single slit diffraction, the condition for the $n^{	ext{th}}$ secondary maxima is given by the formula:
$a \sin 	heta = (2n + 1) \frac{\lambda}{2}$
Here, $a = 0.014 \ mm = 0.014 	imes 10^{-3} \ m$ is the slit width, $	heta = 2.81^{\circ}$ is the angular position, and $n = 2$ for the second bright line.
Rearranging the formula to solve for wavelength $\lambda$:
$\lambda = \frac{2a \sin 	heta}{2n + 1}$
Substituting the given values:
$\lambda = \frac{2 	imes 0.014 	imes 10^{-3} 	imes \sin(2.81^{\circ})}{2(2) + 1}$
$\lambda = \frac{2 	imes 0.014 	imes 10^{-3} 	imes 0.049072}{5}$
$\lambda = \frac{0.028 	imes 10^{-3} 	imes 0.049072}{5}$
$\lambda = 0.0056 	imes 10^{-3} 	imes 0.049072 \approx 2.748 	imes 10^{-7} \ m$
Converting to $\ &Aring;$s $(1 \ &Aring; = 10^{-10} \ m)$:
$\lambda = 2748 \ &Aring;$

$A$ monochromatic light is incident on a single slit of width $0.014 \ mm$. The angular position of the second bright line observed is $2.81^{\circ}$. Then the wavelength of the incident light is $\left[\sin \left(2.81^{\circ}\right)=0.049072\right]$ (in $Å$)

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