Question 251: Waves Optics

Question

In a single slit diffraction pattern, the distance between the plane of the slit and screen is 1.3 m . The width of the slit is 0.65 mm and the second maximum is formed at the distance of 2.6 mm from the centre of the screen. The wavelength of light used is

Accepted Answer

For $\mathrm{n}^{	ext {th }}$ secondary maximum $a \sin 	heta=\left(n+\frac{1}{2}ight) \lambda$
For second maximum, $\mathrm{n}=2$ a $\sin 	heta=\frac{5}{2} \lambda$
For small angles, $\sin 	heta \approx 	an 	heta=\frac{y}{D}$
$$ \begin{aligned} a \frac{y}{D} & =\frac{5}{2} \lambda \ \lambda & =\frac{2 a y}{5 D} \ 	herefore \quad \lambda & =\frac{2 	imes 0.65 	imes 10^{-3} 	imes 2.6 	imes 10^{-3}}{5 	imes 1.3} \ & =0.52 	imes 10^{-6} \ & =5200 \mathop {m{A}}\limits^{m{o}} \end{aligned} $$

Answer

$6500 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$6000 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$5200 \mathop {\rm{A}}\limits^{\rm{o}}$

Answer

$4600\mathop {\rm{A}}\limits^{\rm{o}}$

Waves Optics

In a single slit diffraction pattern, the distance between the plane of the slit and screen is 1.3 m . The width of the slit is 0.65 mm and the second maximum is formed at the distance of 2.6 mm from the centre of the screen. The wavelength of light used is

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