Q44 / 48 · JEE Advanced 2025 (Paper 1) · Physics
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 $m^{\text {th }}$ 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 maximum 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 ______.
Correct answer: 94.5
Solution
Original question asked about absolute error which can be both minimum or maximum, we added word maximum to make it clear for students
$\begin{aligned} & \mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=360 \mu \mathrm{~m} \\ & \mathrm{~b}_{\max }=\frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \mu \mathrm{~m}=454.5 \mu \mathrm{~m} \\ & \mathrm{~b}_{\min }=\frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m}\end{aligned}$
Maximum value of b gives error, $\Delta \mathrm{b}_1=94.5 \mu \mathrm{~m}$
Minimum value of $b$ gives error, $\Delta b_2=63 \mu \mathrm{~m}$
$\therefore$ We always report the largest error, hence correct answer should be $94.5 \mu \mathrm{~m}$
Hence, the correct answer is 94.5.