Name: Position of Maxima in Single Slit Diffraction
Uploaded: 2019-07-19
Duration: 4 min 45 s
Description: Intensities of the bright fringes were equal for the double-slit experiment. Will this also be the case for the single-slit diffraction pattern? Let's find out.

Question

Draw a neat labelled diagram of Young $'$ s Double Slit experiment. Show that

β = \frac{λ}{α}

, where the terms have their usual meaning

(

either for bright or dark fringe

)

.

Accepted Answer

Consider a point $P$ on the viewing screen, the screen is located at a perpendicular distance $β = \frac{λ}{α}$ from the screen containing the slits $S_{1}$ and $S_{2}$ , which are separated by a distance $d$ as shown in figure. Let us assume that the source of light is monochromatic having a wavelength $λ$ . Under these conditions, the waves emerging from $S_{1}$ and $S_{2}$ have the same frequency and amplitude and are in phase.

The light intensity on the screen at $P$ is the resultant of the lights coming from both the slits. As seen in the diagram a wave from the lower slit travels farther than a wave from the upper slit. This difference in path is called the path difference. Therefore, if $' x'$ is the path difference between the waves from slits $S_{2}$ and $S_{1}$ at point $P$ , then

$x = r_{2} - r_{1} = d \sin θ ‐ equation (1)$

where it is assumed that $r_{1}$ and $r_{2}$ are parallel, which is approximately true because $D$ is much greater than $d$ . The value of this path difference determines whether or not the two waves are in phase, when they arrive at point $P$ .

$x = d \sin θ = n λ ‐ equation (2)$

In order to find the position of maxima measured vertically from $O$ to $P$ i.e., $y$ we assume that $D > > d$ . Under these conditions, $θ$ is small, and so the approximation $\sin θ ~ \tan θ$ . Therefore, from $∆ P O Q$ in figure, we find that

$\sin θ \approx \tan θ = \frac{y}{D} ‐ equation (3)$

Substituting this in equation $(2)$ , we have

$d \frac{y}{D} = n λ ‐ equation (4)$

Rewriting the above equation, we have

$y = \frac{D n λ}{d}$

This gives the distance of the $n t h$ maxima from the centre $O$ of the screen. For $n = 0, 1, 2, 3, 4, . . .$ the values of $y$ are

$y_{0} = 0$ $. . .$ position of central bright fringe

$y_{1} = \frac{D λ}{d}$ $. . .$ position of first bright fringe

$y_{2} = \frac{2 D λ}{d}$ $. . .$ position of second bright fringe

$y_{n} = \frac{n D λ}{d}$ $. . .$ position of the $n t h$ bright fringe

The spacing between any two bright fringes gives the width of the dark fringe. Therefore,

$y_{n} - y_{n - 1} = \frac{n D λ}{d} - \frac{(n - 1) D λ}{d} = \frac{D λ}{d}$

This gives the width of the dark fringe.

So the fringe width is given by, $β = \frac{D λ}{d}$

Draw a neat labelled diagram of Young $'$ s Double Slit experiment. Show that $β = \frac{λ}{α}$ , where the terms have their usual meaning $($ either for bright or dark fringe $)$ .

Important Questions on Diffraction of Light

Learn and Prepare for any exam you want !

Draw a neat labelled diagram of Young's Double Slit experiment. Show that β=λα, where the terms have their usual meaning (either for bright or dark fringe).

Important Questions on Diffraction of Light

Learn and Prepare for any exam you want !

Draw a neat labelled diagram of Young $'$ s Double Slit experiment. Show that $β = \frac{λ}{α}$ , where the terms have their usual meaning $($ either for bright or dark fringe $)$ .