A single slit diffraction experiment is performed to determine the slit width using the equation, $$\frac{bd}{D} = m\lambda$$, 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 absolute error (in $$\mu$$m) in the value of $$b$$ estimated using the diffraction maximum that occurs for $$m = 3$$ with $$d = 5$$ mm and $$D = 1$$ m is ________.

The relation for the position of the $$m^{\text{th}}$$ single-slit diffraction maximum is

$$\frac{bd}{D}=m\lambda \;\Longrightarrow\; b=\frac{m\lambda D}{d}$$

Given data:
$$m=3,\; \lambda = 600\,\text{nm}=600\times 10^{-9}\,\text{m},\; D=1\,\text{m},\; d=5\,\text{mm}=0.005\,\text{m}$$

Substituting,

$$b=\frac{3\,(600\times 10^{-9})\,(1)}{0.005}=3.6\times 10^{-4}\,\text{m}=360\,\mu\text{m}$$

The least counts (LC) of the measuring scales are
LC for $$D$$ : $$1\,\text{cm}=0.01\,\text{m}\;\Longrightarrow\; \Delta D = 0.01\,\text{m}$$
LC for $$d$$ : $$1\,\text{mm}=0.001\,\text{m}\;\Longrightarrow\; \Delta d = 0.001\,\text{m}$$

For a quantity obtained as a product/quotient, the maximum fractional error adds algebraically:

$$\frac{\Delta b}{b}= \frac{\Delta D}{D}+\frac{\Delta d}{d}$$

Compute each term:
$$\frac{\Delta D}{D}= \frac{0.01}{1}=0.01$$
$$\frac{\Delta d}{d}= \frac{0.001}{0.005}=0.20$$

Hence
$$\frac{\Delta b}{b}=0.01+0.20 = 0.21$$

Absolute error in $$b$$:

$$\Delta b = b \left(0.21\right)=\left(3.6\times 10^{-4}\right)(0.21)=7.56\times 10^{-5}\,\text{m}$$

Convert to micrometres ($$1\,\text{m}=10^{6}\,\mu\text{m}$$):

$$\Delta b = 7.56\times 10^{-5}\times 10^{6}=75.6\,\mu\text{m}$$

Therefore, the absolute error in the estimated slit width lies in the range $$75\text{-}79\,\mu\text{m}$$.