Question 323: Waves Optics

Question

Light of wavelength $$5000 \mathop A\limits^o$$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of $$5 \mathrm{~mm}$$ from the central maximum on a screen placed at a distance of $$2 \mathrm{~m}$$ from the slit. The width of the slit is

Accepted Answer

Position of the $$n$$th minima is
$$x_n=\frac{n \lambda D}{a}$$ .... (i)
where, $$a=$$ slit width
$$D=$$ Distance between screen
and slit
$$\lambda=$$ wavelength
Given, $$\lambda=5000 \mathop A\limits^o=5000 	imes 10^{-10} \mathrm{~m}$$
$$\begin{aligned}
x & =5 \mathrm{~mm}=5 	imes 10^{-3} \mathrm{~m} \
\quad D & =2 \mathrm{~m} \
\Rightarrow \quad n & =1
\end{aligned}$$
Put all these values in Eq. (i), we get
$$\begin{array}{rlrl}
& 5 	imes 10^{-3} =\frac{1 	imes 5000 	imes 10^{-10} 	imes 2}{a} \
\Rightarrow & a =\frac{5 	imes 10^{-7} 	imes 2}{5 	imes 10^{-3}} \
\Rightarrow & a =2 	imes 10^{-4} \mathrm{~m} \
& =0.02 \mathrm{~cm}
\end{array}$$
Therefore, the slit width is $$0.02 \mathrm{~cm}$$.

Answer

2 cm

Answer

0.2 cm

Answer

0.02 cm

Answer

0.01 cm

Waves Optics

Light of wavelength $5000 \mathop A\limits^o is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of 5 \mathrm{~mm} from the central maximum on a screen placed at a distance of 2 \mathrm{~m}$ from the slit. The width of the slit is

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