Question 48: Waves Optics

Question

A single slit diffraction experiment is performed to determine the slit width using the equation, $\frac{b d}{D}=$ $m \lambda$, where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m^{	ext {th }}$ diffraction maximum and the central maximum, and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of 1 cm and 1 mm , respectively. The values of $\lambda$ and $m$ are known precisely to be 600 nm and 3, respectively. The absolute error (in $\mu \mathrm{m}$ ) in the value of $b$ estimated using the diffraction maximum that occurs for $m=3$ with $d=5 \mathrm{~mm}$ and $D=1 \mathrm{~m}$ is ________.

Accepted Answer

Given. $\frac{b d}{d}=m \lambda$
so $\quad b=\frac{m \lambda D}{d}$
Error propagation:
$$ \frac{\Delta b}{b}=\frac{\Delta m}{m}+\frac{\Delta \lambda}{\lambda}+\frac{\Delta D}{D}+\frac{\Delta d}{d} $$
as $\lambda $ and $ m$ are known precisely so $\Delta \lambda$ and $\Delta$m are zero.
So ${\Delta }{b}=\frac{\Delta D}{D}+\frac{\Delta d}{d}$
$$ \begin{aligned} & \Rightarrow {\Delta }{b}=\left(\frac{\Delta D}{D}+\frac{\Delta d}{d}ight)\left(\frac{m \lambda D}{d}ight) \ & \Rightarrow  {\Delta }{b} =\left(\frac{1}{100}+\frac{0.1}{0.5}ight)\left(\frac{3 	imes 600 	imes 10^{} 	imes 100}{0.5}ight) \ &=\left(\frac{1+20}{100}ight)\left(\frac{18}{5} 	imes \frac{1}{100}ight) \ &=\frac{21}{100} 	imes \frac{18}{5} 	imes \frac{1}{100} \mathrm{~cm} \ & \end{aligned} $$
$$ \begin{aligned} &\begin{gathered} \Delta b=\frac{1890}{25000} 	imes 10 \mu\mathrm{m}=\frac{1890}{25} \mu\mathrm{m} \ \Rightarrow \Delta b=75.60 \mu\mathrm{m} \end{gathered}\ &	ext {  } \end{aligned} $$

Waves Optics

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