Question 199: Wave Optics

Question

In a single slit diffraction pattern, a light of wavelength 6000$$\mathop A\limits^o$$ is used. The distance between the first and third minima in the diffraction pattern is found to be $$3 \mathrm{~mm}$$ when the screen in placed $$50 \mathrm{~cm}$$ away from slits. The width of the slit is _________ $$	imes 10^{-4} \mathrm{~m}$$.

Accepted Answer

For $$\mathrm{n}^{	ext {th }}$$ minima
$$\mathrm{b} \sin 	heta=\mathrm{n} \lambda$$
($$\lambda$$ is small so $$\sin 	heta$$ is small, hence $$\sin 	heta \simeq 	an 	heta$$)
$$\mathrm{btan} 	heta=\mathrm{n} \lambda$$
$$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$$
$$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(	ext { Position of } \mathrm{n}^{\mathrm{th}}ight. 	ext { minima) }$$
$$\begin{aligned}
& \mathrm{B} ightarrow 1^{	ext {st }} 	ext { minima, } \mathrm{A} ightarrow 3^{\mathrm{rd}} 	ext { minima } \
& \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \
& \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \
& 3 	imes 10^{-3}=\frac{2 	imes 6000 	imes 10^{-10} 	imes 0.5}{\mathrm{~b}} \
& \mathrm{~b}=\frac{2 	imes 6000 	imes 10^{-10} 	imes 0.5}{3 	imes 10^{-3}} \
& \mathrm{~b}=2 	imes 10^{-4} \mathrm{~m} \
& x=2
\end{aligned}$$

Wave Optics

Integer / Numeric Answer Type