A single slit diffraction experiment is performed to determine the slit width using the equation, $\

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 $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

Alternate method: Error in $b$ is too large. We may consider solution for the given problem as, $$ \begin{gathered} \quad b=3.6 	imes 10^{-4} \mathrm{~m} \ b_{\max }=\frac{3 	imes 600 	imes 10^{-3} 	imes 1.01}{4 	imes 10^{-3}} \mu \mathrm{~m}=454.5 \mu \mathrm{~m} \ b_{\min }=\frac{3 	imes 600 	imes 10^{-3} 	imes 0.99}{6 	imes 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m} \ 	herefore \quad \Delta b_{1}=94.5 \mu \mathrm{~m}(	ext { For max. value of } b) \end{gathered} $$