Question 12

The slit width is given by $b = \frac{m \lambda D}{d}$. Taking natural log on both sides: $\ln b = \ln m + \ln \lambda + \ln D - \ln d$. Differentiating both sides to find relative error (assuming $m$ and $\lambda$ are constants with zero error): $\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d}$. Given values: $m = 3$, $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$, $d = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}$, $D = 1 \text{ m}$. Least counts (errors): $\Delta D = 1 \text{ cm} = 0.01 \text{ m}$, $\Delta d = 1 \text{ mm} = 0.001 \text{ m}$. First, calculate $b$: $b = \frac{3 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-3}} = \frac{1800 \times 10^{-9}}{5 \times 10^{-3}} = 360 \times 10^{-6} \text{ m} = 360 \mu\text{m}$. Now, calculate the absolute error $\Delta b$: $\Delta b = b \left( \frac{\Delta D}{D} + \frac{\Delta d}{d} \right)$ $\Delta b = 360 \left( \frac{0.01}{1} + \frac{0.001}{0.005} \right)$ $\Delta b = 360 (0.01 + 0.2) = 360 \times 0.21 = 75.6 \mu\text{m}$. The value lies within the range [75, 79].