Question

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

Hint:

When using a formula involving multiple measured quantities, use relative error addition to estimate the error in the result: \( \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} + \dots \) for multiplication/division.

Answer 1

Correct Answer: 75.6

To solve for the absolute error in the slit width \( b \) (in \(\mu m\)), we start with the given equation for single slit diffraction: \[ \frac{bd}{D} = m\lambda \] Step 1: Solve for \( b \) The equation relates the slit width \( b \), the distance \( d \) between the \( m \)-th diffraction maximum and the central maximum, the distance \( D \) between the slit and the screen, and the wavelength \( \lambda \). Rearranging to isolate \( b \): \[ b = \frac{m\lambda D}{d} \] Given:

• \( m = 3 \),
• \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} = 6 \times 10^{-7} \, \text{m} \),
• \( D = 1 \, \text{m} \),
• \( d = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \).

Substitute the values: \[ b = \frac{3 \times (6 \times 10^{-7}) \times 1}{5 \times 10^{-3}} = \frac{18 \times 10^{-7}}{5 \times 10^{-3}} = \frac{18}{5} \times 10^{-4} = 3.6 \times 10^{-4} \, \text{m} \] Convert \( b \) to micrometers (\( 1 \, \text{m} = 10^6 \, \mu m \)): \[ b = 3.6 \times 10^{-4} \times 10^6 = 360 \, \mu m \] Step 2: Calculate the absolute error in \( b \) The absolute error in \( b \) arises from the measurement errors in \( d \), \( D \), \( \lambda \), and \( m \). The problem states that \( D \) and \( d \) are measured with scales of least count \( 1 \, \text{cm} \) and \( 1 \, \text{mm} \), respectively, and \( \lambda \) and \( m \) are known precisely.

• Least count of \( D \): \( \Delta D = 1 \, \text{cm} = 0.01 \, \text{m} \),
• Least count of \( d \): \( \Delta d = 1 \, \text{mm} = 0.001 \, \text{m} \),
• \( \lambda \) and \( m \): no error.

Using the formula for propagation of relative error: \[ \frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d} \] \[ \frac{\Delta D}{D} = \frac{0.01}{1} = 0.01, \quad \frac{\Delta d}{d} = \frac{0.001}{0.005} = 0.2 \] \[ \frac{\Delta b}{b} = 0.01 + 0.2 = 0.21 \] Now calculate the absolute error: \[ \Delta b = b \times \frac{\Delta b}{b} = 360 \times 0.21 = 75.6 \, \mu m \] Final Answer: The absolute error in the value of \( b \) is: \[ \boxed{75.6 \, \mu m} \]