The Fraunhofer diffraction pattern of a singl | vedclass

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(A) For a single slit diffraction, the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$.Given that the angle $	heta$ is ver

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$5000$

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(A) For a single slit diffraction, the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$.Given that the angle $	heta$ is very small, we can approximate $\sin 	heta \approx 	an 	heta = \frac{x}{f}$, where $x$ is the distance from the central maximum and $f$ is the focal length.Substituting this into the condition for the $n^{th}$ minimum: $a \left( \frac{x}{f} ight) = n \lambda$.For the third minimum, $n = 3$. Given $a = 0.3 	imes 10^{-3} \, m$, $x = 5 	imes 10^{-3} \, m$, and $f = 1 \, m$.Rearranging for $\lambda$: $\lambda = \frac{a x}{3 f}$.$\lambda = \frac{0.3 	imes 10^{-3} 	imes 5 	imes 10^{-3}}{3 	imes 1} = \frac{1.5 	imes 10^{-6}}{3} = 0.5 	imes 10^{-6} \, m$.Converting to $\mathring{A}$: $\lambda = 0.5 	imes 10^{-6} 	imes 10^{10} \, \mathring{A} = 5000 \, \mathring{A}$.

The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length $1\,m$. The width of the slit is $0.3\,mm$. If the third minimum is formed at a distance of $5\,mm$ from the central maximum, then the wavelength of light will be......$\mathring{A}$

Similar Questions

Orange light of wavelength $6000 \times 10^{-10} \ m$ illuminates a single slit of width $0.6 \times 10^{-4} \ m$. The maximum possible number of diffraction minima produced on both sides of the central maximum is $........$

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