Question

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 $m^{th}$ 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 ___.

Answer 1

37.8

Answer 2

The given equation relating the slit width $b$, the distance $d$ of the $m^{th}$ diffraction maximum from the central maximum, the distance $D$ between the slit and the screen, the order $m$, and the wavelength $\lambda$ is $\frac{bd}{D} = m\lambda$.

We need to find the absolute error in the value of $b$. From the given equation, we can express $b$ as: $b = \frac{m\lambda D}{d}$

We are given the following values: $m = 3$ (known precisely, $\Delta m = 0$) $\lambda = 600$ nm $= 600 \times 10^{-9}$ m (known precisely, $\Delta \lambda = 0$) $d = 5$ mm $= 5 \times 10^{-3}$ m $D = 1$ m

The errors in the measurements of $D$ and $d$ are related to the least count of the scales used. Least count for $D$ is 1 cm $= 10^{-2}$ m. The absolute error in $D$ is $\Delta D = \frac{1}{2} \times \text{least count} = \frac{1}{2} \times 10^{-2}$ m $= 0.5 \times 10^{-2}$ m. Least count for $d$ is 1 mm $= 10^{-3}$ m. The absolute error in $d$ is $\Delta d = \frac{1}{2} \times \text{least count} = \frac{1}{2} \times 10^{-3}$ m $= 0.5 \times 10^{-3}$ m.

To find the absolute error in $b$, we use the formula for propagation of errors. The formula for $b$ is $b = \frac{m\lambda D}{d}$. Since $m$ and $\lambda$ are known precisely, they act as constants in the error calculation. Let $C = m\lambda$. Then $b = C \frac{D}{d}$.

The maximum relative error in $b$ is given by the sum of the relative errors in the measured quantities $D$ and $d$: $\frac{\Delta b}{b} = \left| \frac{\Delta C}{C} \right| + \left| \frac{\Delta D}{D} \right| + \left| \frac{\Delta d}{d} \right|$ Since $C = m\lambda$ has no error, $\Delta C = 0$. $\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d}$

First, calculate the value of $b$ using the given values: $b = \frac{3 \times (600 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{5 \times 10^{-3} \text{ m}} = \frac{1800 \times 10^{-9}}{5 \times 10^{-3}} \text{ m} = \frac{1.8 \times 10^{-6}}{5 \times 10^{-3}} \text{ m} = 0.36 \times 10^{-3} \text{ m}$ $b = 3.6 \times 10^{-4} \text{ m}$

Now, calculate the relative errors in $D$ and $d$: $\frac{\Delta D}{D} = \frac{0.5 \times 10^{-2} \text{ m}}{1 \text{ m}} = 0.5 \times 10^{-2} = 0.005$ $\frac{\Delta d}{d} = \frac{0.5 \times 10^{-3} \text{ m}}{5 \times 10^{-3} \text{ m}} = \frac{0.5}{5} = 0.1$

The relative error in $b$ is: $\frac{\Delta b}{b} = 0.005 + 0.1 = 0.105$

Now, calculate the absolute error in $b$: $\Delta b = b \times \left( \frac{\Delta D}{D} + \frac{\Delta d}{d} \right)$ $\Delta b = (3.6 \times 10^{-4} \text{ m}) \times (0.105)$ $\Delta b = 0.378 \times 10^{-4} \text{ m}$

The question asks for the absolute error in $\mu$m. $1 \text{ m} = 10^6 \ \mu\text{m}$ $\Delta b = 0.378 \times 10^{-4} \times 10^6 \ \mu\text{m} = 0.378 \times 10^2 \ \mu\text{m} = 37.8 \ \mu\text{m}$.

The final answer is $\boxed{37.8}$.