Single Slit Diffraction

To find the width of the slit in a single slit diffraction pattern, we can use the formula for the position of minima: $a%20%5Csin%20%5Ctheta%20%3D%20n%5Clambda$ , where $a$ is the slit width, $%5Ctheta$ is the diffraction angle, $n$ is the order of the minima, and $%5Clambda$ is the wavelength of light. For small angles, $%5Csin%20%5Ctheta%20%5Capprox%20%5Ctan%20%5Ctheta%20%3D%20%5Cfrac%7By%7D%7BD%7D$ , where $y$ is the distance of the minima from the central maximum on the screen, and $D$ is the distance from the slit to the screen.

The distance between the first ( $n%3D1$ ) and third ( $n%3D3$ ) minima is given as 3 mm. Using the positions of the minima: $y_1%20%3D%20%5Cfrac%7B%5Clambda%20D%7D%7Ba%7D$ and $y_3%20%3D%20%5Cfrac%7B3%5Clambda%20D%7D%7Ba%7D$ .

The difference is: $y_3%20-%20y_1%20%3D%20%5Cfrac%7B3%5Clambda%20D%7D%7Ba%7D%20-%20%5Cfrac%7B%5Clambda%20D%7D%7Ba%7D%20%3D%20%5Cfrac%7B2%5Clambda%20D%7D%7Ba%7D$ .

Equating this to the given distance: $%5Cfrac%7B2%5Clambda%20D%7D%7Ba%7D%20%3D%203%20%5C%2C%20%5Ctext%7Bmm%7D%20%3D%200.003%20%5C%2C%20%5Ctext%7Bm%7D$ .

Substitute $%5Clambda%20%3D%206000%20%5C%2C%20%5Ctext%7B%C3%85%7D%20%3D%206000%20%5Ctimes%2010%5E%7B-10%7D%20%5C%2C%20%5Ctext%7Bm%7D$ and $D%20%3D%2050%20%5C%2C%20%5Ctext%7Bcm%7D%20%3D%200.5%20%5C%2C%20%5Ctext%7Bm%7D$ :

$a%20%3D%20%5Cfrac%7B2%20%5Ccdot%206000%20%5Ctimes%2010%5E%7B-10%7D%20%5Ccdot%200.5%7D%7B0.003%7D$ .

$a%20%3D%20%5Cfrac%7B6000%20%5Ctimes%2010%5E%7B-10%7D%7D%7B0.003%7D%20%3D%202%20%5Ctimes%2010%5E%7B-4%7D%20%5C%2C%20%5Ctext%7Bm%7D$ .

Thus, the width of the slit is $2%20%5Ctimes%2010%5E%7B-4%7D%20%5C%2C%20%5Ctext%7Bm%7D$ , fitting well within the specified range [2,2].

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Single Slit Diffraction

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs