Error Propagation

A single slit diffraction experiment is performed to determine the slit width using the equation, $%5Cdfrac%7Bb%20d%7D%7BD%7D%20%3D%20m%20%5Clambda$ , where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m%5E%7B%5Ctext%7Bth%7D%7D$ diffraction maximum and the central maximum, and $%5Clambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of 1 cm and 1 mm, respectively. The values of $%5Clambda$ and $m$ are known precisely to be $600%20%5C%2C%20%5Ctext%7Bnm%7D$ and 3, respectively. The absolute error (in $%5Cmu%5Ctext%7Bm%7D$ ) in the value of $b$ estimated using the diffraction maximum that occurs for $m%20%3D%203$ with $d%20%3D%205%20%5C%2C%20%5Ctext%7Bmm%7D$ and $D%20%3D%201%20%5C%2C%20%5Ctext%7Bm%7D$ is ___.

To solve for the absolute error in the slit width

b

(in

%5Cmu%20m

), we start with the given equation for single slit diffraction:

%5Cfrac%7Bbd%7D%7BD%7D%20%3D%20m%5Clambda

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

%5Clambda

. Rearranging to isolate

b

b%20%3D%20%5Cfrac%7Bm%5Clambda%20D%7D%7Bd%7D

Given:

$m%20%3D%203$ ,
$%5Clambda%20%3D%20600%20%5C%2C%20%5Ctext%7Bnm%7D%20%3D%20600%20%5Ctimes%2010%5E%7B-9%7D%20%5C%2C%20%5Ctext%7Bm%7D%20%3D%206%20%5Ctimes%2010%5E%7B-7%7D%20%5C%2C%20%5Ctext%7Bm%7D$ ,
$D%20%3D%201%20%5C%2C%20%5Ctext%7Bm%7D$ ,
$d%20%3D%205%20%5C%2C%20%5Ctext%7Bmm%7D%20%3D%205%20%5Ctimes%2010%5E%7B-3%7D%20%5C%2C%20%5Ctext%7Bm%7D$ .

Substitute the values:

b%20%3D%20%5Cfrac%7B3%20%5Ctimes%20(6%20%5Ctimes%2010%5E%7B-7%7D)%20%5Ctimes%201%7D%7B5%20%5Ctimes%2010%5E%7B-3%7D%7D%20%3D%20%5Cfrac%7B18%20%5Ctimes%2010%5E%7B-7%7D%7D%7B5%20%5Ctimes%2010%5E%7B-3%7D%7D%20%3D%20%5Cfrac%7B18%7D%7B5%7D%20%5Ctimes%2010%5E%7B-4%7D%20%3D%203.6%20%5Ctimes%2010%5E%7B-4%7D%20%5C%2C%20%5Ctext%7Bm%7D

Convert

b

to micrometers (

1%20%5C%2C%20%5Ctext%7Bm%7D%20%3D%2010%5E6%20%5C%2C%20%5Cmu%20m

b%20%3D%203.6%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%2010%5E6%20%3D%20360%20%5C%2C%20%5Cmu%20m

Step 2: Calculate the absolute error in $b$ The absolute error in

b

arises from the measurement errors in

d

D

%5Clambda

, and

m

. The problem states that

D

and

d

are measured with scales of least count

1%20%5C%2C%20%5Ctext%7Bcm%7D

and

1%20%5C%2C%20%5Ctext%7Bmm%7D

, respectively, and

%5Clambda

and

m

are known precisely.

Least count of $D$ : $%5CDelta%20D%20%3D%201%20%5C%2C%20%5Ctext%7Bcm%7D%20%3D%200.01%20%5C%2C%20%5Ctext%7Bm%7D$ ,
Least count of $d$ : $%5CDelta%20d%20%3D%201%20%5C%2C%20%5Ctext%7Bmm%7D%20%3D%200.001%20%5C%2C%20%5Ctext%7Bm%7D$ ,
$%5Clambda$ and $m$ : no error.

Using the formula for propagation of relative error:

%5Cfrac%7B%5CDelta%20b%7D%7Bb%7D%20%3D%20%5Cfrac%7B%5CDelta%20D%7D%7BD%7D%20%2B%20%5Cfrac%7B%5CDelta%20d%7D%7Bd%7D

%5Cfrac%7B%5CDelta%20D%7D%7BD%7D%20%3D%20%5Cfrac%7B0.01%7D%7B1%7D%20%3D%200.01%2C%20%5Cquad%20%5Cfrac%7B%5CDelta%20d%7D%7Bd%7D%20%3D%20%5Cfrac%7B0.001%7D%7B0.005%7D%20%3D%200.2

%5Cfrac%7B%5CDelta%20b%7D%7Bb%7D%20%3D%200.01%20%2B%200.2%20%3D%200.21

Now calculate the absolute error:

%5CDelta%20b%20%3D%20b%20%5Ctimes%20%5Cfrac%7B%5CDelta%20b%7D%7Bb%7D%20%3D%20360%20%5Ctimes%200.21%20%3D%2075.6%20%5C%2C%20%5Cmu%20m

Final Answer: The absolute error in the value of

b

is:

%5Cboxed%7B75.6%20%5C%2C%20%5Cmu%20m%7D

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Error Propagation

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs