# Young’s Double Slit Experiment

Thomas Young in 1801 carried out an experiment to study the behaviour of light in microscopic level. This experiment is commonly known as Young’s double slit experiment. In this experiment, a monochromatic light is passed through the slit and the behaviour of light is studied.

Let us consider a monochromatic light S is passed through two pinholes ‘S_{1}‘ and ‘S_{2}‘ which are two coherent sources emitted from single point ‘S’. The point C is equidistance from S_{1} and S_{2}. Therefore, path difference at C is zero. Let D is the distance between the slits and screen and d is the distance between two slits.

Let us consider a point P is at distance x from C. The path difference between two waves arriving at P is given by:

Path difference = S_{2}P =S_{1}P

By using simple geometrical laws, we can have the following relations:

$$ \text{PQ} = x – \frac{d}{2}$$

$$ \text{and PR} = x + \frac{d}{2}$$

$$\text{S}_1\text{P} =\text{D – PQ}……(i)$$

$$\text{S}_2\text{P} =\text{D + PR}……(ii)$$

Squaring and subtracting equation (i) from (ii),

$${(\text{S}_2\text{P})}^2 – {(\text{S}_1\text{P})^2} = \left [ \text{D}^2 -\left (x + \frac{d}{2} \right )^2\right ] – \left [ \text{D}^2 + \left (x -\frac{d}{2} \right )^2 \right ]$$

$$\text{or, (S}_2\text{P} – \text{S}_1\text{P}) {(S}_2\text{P} + \text{S}_1\text{P}) = \text{2xd}$$

$$\text{or, (S}_2\text{P} – \text{S}_1\text{P}) = \frac{\text{2xd}}{\text{BP + AP}}$$

Since the point P lies very close to C,

$$\text{S}_2\text{P} \approx \text{S}_1\text{P} \approx \text{D}$$

$$\text{So,} \text{S}_1\text{P} + \text{S}_2\text{P} = \text{D + D = 2D}$$

Now, from path difference relation:

$$ \therefore \text{Path difference} = \text{S}_2\text{P} – \text{S}_1\text{P} = \frac{\text{2xd}}{2D} = \frac{\text{xd}}{\text{D}}$$

**Bright fringes**

When path difference = n λ, where n = 0, 1, 2, 3, 4, 5 etc., then constructive interference or bright fringes are seen.

$$So, \frac{\text{xd}}{\text{D}} = \text{n}\lambda $$

$$ \text{or, x}_{\text{n}} = \frac{\text{n}\lambda \text{D}}{\text{d}}$$

Here, n represents the order of fringes. This gives the distance of nth bright fringe from the centre C.

When n = 0, then x_{o} = 0 = central maxima

When n = 1, then x_{1} = λD/d = first bright fringe

When n = 2, then x_{2} = 2λD/d = second bright fringe

Here, the distance between two consecutive bright fringes is called fringe width.

$$\text{Now, Fringe width}(\beta) = \text{x}_2 – \text{x}_1 = \frac{\text{2}\lambda \text{D}}{\text{d}} – \frac{\lambda \text{D}}{\text{d}} = \frac{\lambda \text{D}}{\text{d}}$$

**Dark fringes**

When path difference = (2n-1) λ/2, where n = 0, 1, 2, 3, 4, 5 etc., then constructive interference or bright fringes are seen.

$$So, \frac{\text{xd}}{\text{D}} = \text({2n-1})\frac{\lambda }{2} $$

$$ \text{or, x}_{\text{n}} = \frac{(\text{2n + 1})\lambda \text{D}}{\text{2d}}$$

The above expression gives the distance of nth dark fringe from the centre C.

When n = 1, x_{1} = λD/2d = first dark fringe

When n = 2, x_{1} = 3λD/2d = second dark fringe

Here, the distance between two consecutive dark fringes is called fringe width.

$$\text{So, Fringe width}(\beta) = \text{x}_2 – \text{x}_1 = \frac{\text{3}\lambda \text{D}}{\text{d}} – \frac{\lambda \text{D}}{\text{d}}$$

Since, width of bright fringe and dark fringe is equal, so we can write:

$$\beta = \frac{\lambda \text{D}}{\text{d}}$$

So from the above expression, we understand that when fringe width increases, the wavelength also increases, the distance D of the screen from the source increases and distance d between the source decreases.

Do you like this article ? If yes then like otherwise dislike : 6 0

## No Responses to “Young’s Double Slit Experiment”