Given relation

The slit width is estimated from $\frac{bd}{D}=m\lambda$ so $b=\frac{m\lambda D}{d}.$

Given data

$m=3,\quad \lambda=600\,\text{nm}=600\times10^{-9}\,\text{m}$ $D=1\,\text{m},\quad d=5\,\text{mm}=5\times10^{-3}\,\text{m}$

Least counts:

• for $D$: $1\,\text{cm}=10^{-2}\,\text{m}$
• for $d$: $1\,\text{mm}=10^{-3}\,\text{m}$

Since $m$ and $\lambda$ are known precisely, their errors are zero.

Compute slit width

$b=\frac{3\times 600\times10^{-9}\times 1}{5\times10^{-3}}$

$b=\frac{1800\times10^{-9}}{5\times10^{-3}}=3.6\times10^{-4}\,\text{m}$

$b=360\,\mu\text{m}.$

Error propagation

Since $b\propto \frac{D}{d},$ the maximum fractional error is $\frac{\Delta b}{b}=\frac{\Delta D}{D}+\frac{\Delta d}{d}.$

Taking absolute errors equal to the least counts: $\Delta D=1\,\text{cm}=0.01\,\text{m},\qquad \Delta d=1\,\text{mm}=0.001\,\text{m}.$

Therefore, $\frac{\Delta b}{b}=\frac{0.01}{1}+\frac{0.001}{0.005}=0.01+0.2=0.21.$

Absolute error in $b$

$\Delta b=0.21\times 360\,\mu\text{m}=75.6\,\mu\text{m}.$

For integer answer, this is approximately $\boxed{76}.$

Comparison with stored answer

Stored correct answer is 75TO79, i.e. acceptable range $75$ to $79$.

Our derived answer $76$ lies in this range, so it agrees.