Interference

                                         screen
                                           |
                                           P   ^
              S1 •                         |   | y
          d  ----+------------------------ O'  v
              S2 •                         |
                 |<--------- D --------->|
     (S1P and S2P are the two paths to P; O' is the centre)

Setup: $S_1$ and $S_2$ are two coherent narrow slits a distance $d$ apart, lit by monochromatic light of wavelength $\lambda$. The screen is at distance $D$ ($D \gg d$). $O'$ is the centre of the screen and $P$ is a point at distance $y$ from $O'$.

1. By Pythagoras' theorem,

$$(S_2P)^2 = D^2 + \left(y + \frac{d}{2}\right)^2, \qquad (S_1P)^2 = D^2 + \left(y - \frac{d}{2}\right)^2$$

2. Subtracting,

$$(S_2P)^2 - (S_1P)^2 = 2yd$$ $$(S_2P - S_1P)(S_2P + S_1P) = 2yd$$

3. Since $D \gg d, y$, we have $S_2P + S_1P \approx 2D$. So the path difference is

$$\Delta = S_2P - S_1P = \frac{yd}{D}$$

4. Bright fringes (constructive interference) occur when $\Delta = n\lambda$:

$$\frac{y_n d}{D} = n\lambda \;\Rightarrow\; y_n = \frac{n\lambda D}{d}, \quad n = 0, 1, 2, \dots$$

Width of a dark fringe = distance between two consecutive bright fringes:

$$\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$

5. Dark fringes (destructive interference) occur when $\Delta = (2n+1)\dfrac{\lambda}{2}$:

$$y_n' = \frac{(2n+1)\lambda D}{2d}$$

Width of a bright fringe = distance between two consecutive dark fringes:

$$\beta' = y_{n+1}' - y_n' = \frac{(2n+3)\lambda D}{2d} - \frac{(2n+1)\lambda D}{2d} = \frac{\lambda D}{d}$$

6. Therefore

$$\boxed{\beta = \beta' = \frac{\lambda D}{d}}$$

The fringe width does not depend on $n$, so bright and dark fringes are of equal width and are equally spaced.

(a) Conditions for sustained interference

1. The two sources must be coherent, i.e. have the same frequency (wavelength) and a constant phase difference.
2. The two waves should have the same (or nearly the same) amplitude, so that dark fringes are really dark.
3. The two sources should be narrow and close together ($d$ small), and the screen should be reasonably far away, so that fringes are wide enough to see.
4. The waves should be in the same state of polarization and travel along nearly the same direction.

(b) Bright and dark fringes are equally spaced

                                         screen
                                           |
                                           P   ^
              S1 •                         |   | y
          d  ----+------------------------ O'  v
              S2 •                         |
                 |<--------- D --------->|
     (S1P and S2P are the two paths to P; O' is the centre)

Setup: $S_1$ and $S_2$ are two coherent narrow slits a distance $d$ apart, lit by monochromatic light of wavelength $\lambda$. The screen is at distance $D$ ($D \gg d$). $O'$ is the centre of the screen and $P$ is a point at distance $y$ from $O'$.

1. By Pythagoras' theorem,

$$(S_2P)^2 = D^2 + \left(y + \frac{d}{2}\right)^2, \qquad (S_1P)^2 = D^2 + \left(y - \frac{d}{2}\right)^2$$

2. Subtracting,

$$(S_2P)^2 - (S_1P)^2 = 2yd$$ $$(S_2P - S_1P)(S_2P + S_1P) = 2yd$$

3. Since $D \gg d, y$, we have $S_2P + S_1P \approx 2D$. So the path difference is

$$\Delta = S_2P - S_1P = \frac{yd}{D}$$

4. Bright fringes (constructive interference) occur when $\Delta = n\lambda$:

$$\frac{y_n d}{D} = n\lambda \;\Rightarrow\; y_n = \frac{n\lambda D}{d}, \quad n = 0, 1, 2, \dots$$

Width of a dark fringe = distance between two consecutive bright fringes:

$$\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$

5. Dark fringes (destructive interference) occur when $\Delta = (2n+1)\dfrac{\lambda}{2}$:

$$y_n' = \frac{(2n+1)\lambda D}{2d}$$

Width of a bright fringe = distance between two consecutive dark fringes:

$$\beta' = y_{n+1}' - y_n' = \frac{(2n+3)\lambda D}{2d} - \frac{(2n+1)\lambda D}{2d} = \frac{\lambda D}{d}$$

6. Therefore

$$\boxed{\beta = \beta' = \frac{\lambda D}{d}}$$

The fringe width does not depend on $n$, so bright and dark fringes are of equal width and are equally spaced.

(c) Numerical

Given: distance from 1st to 5th fringe $= 2.5$ mm, $\lambda = 620$ nm $= 6.2\times10^{-7}$ m, $D = 80$ cm $= 0.80$ m.

From the 1st fringe to the 5th fringe there are 4 fringe widths:

$$4\beta = 2.5 \text{ mm} \;\Rightarrow\; \beta = 0.625 \text{ mm} = 6.25\times10^{-4} \text{ m}$$

Using $\beta = \dfrac{\lambda D}{d}$:

$$d = \frac{\lambda D}{\beta} = \frac{6.2\times10^{-7} \times 0.80}{6.25\times10^{-4}} = 7.94\times10^{-4} \text{ m}$$

Slit separation $\approx$ 0.79 mm.

(a) Coherent sources: Two sources of light are coherent if they emit waves of the same frequency (wavelength) with a constant phase difference (zero or constant) between them. In YDSE, $S_1$ and $S_2$ are coherent because both are lit by the same single slit.

(b) Interference of light: The modification (redistribution) of light intensity in the region where two coherent light waves superpose, giving alternate bright and dark bands (fringes), is called interference of light.

(c) Fringe width

                                         screen
                                           |
                                           P   ^
              S1 •                         |   | y
          d  ----+------------------------ O'  v
              S2 •                         |
                 |<--------- D --------->|
     (S1P and S2P are the two paths to P; O' is the centre)

Setup: $S_1$ and $S_2$ are two coherent narrow slits a distance $d$ apart, lit by monochromatic light of wavelength $\lambda$. The screen is at distance $D$ ($D \gg d$). $O'$ is the centre of the screen and $P$ is a point at distance $y$ from $O'$.

1. By Pythagoras' theorem,

$$(S_2P)^2 = D^2 + \left(y + \frac{d}{2}\right)^2, \qquad (S_1P)^2 = D^2 + \left(y - \frac{d}{2}\right)^2$$

2. Subtracting,

$$(S_2P)^2 - (S_1P)^2 = 2yd$$ $$(S_2P - S_1P)(S_2P + S_1P) = 2yd$$

3. Since $D \gg d, y$, we have $S_2P + S_1P \approx 2D$. So the path difference is

$$\Delta = S_2P - S_1P = \frac{yd}{D}$$

4. Bright fringes (constructive interference) occur when $\Delta = n\lambda$:

$$\frac{y_n d}{D} = n\lambda \;\Rightarrow\; y_n = \frac{n\lambda D}{d}, \quad n = 0, 1, 2, \dots$$

Width of a dark fringe = distance between two consecutive bright fringes:

$$\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$

5. Dark fringes (destructive interference) occur when $\Delta = (2n+1)\dfrac{\lambda}{2}$:

$$y_n' = \frac{(2n+1)\lambda D}{2d}$$

Width of a bright fringe = distance between two consecutive dark fringes:

$$\beta' = y_{n+1}' - y_n' = \frac{(2n+3)\lambda D}{2d} - \frac{(2n+1)\lambda D}{2d} = \frac{\lambda D}{d}$$

6. Therefore

$$\boxed{\beta = \beta' = \frac{\lambda D}{d}}$$

The fringe width does not depend on $n$, so bright and dark fringes are of equal width and are equally spaced.

(d) $\beta = \dfrac{\lambda D}{d}$. With $D' = 2D$ and $d' = d/2$:

$$\beta' = \frac{\lambda (2D)}{d/2} = 4\frac{\lambda D}{d} = 4\beta$$

The fringe width becomes four times.

(e) In water, the wavelength becomes $\lambda_w = \dfrac{\lambda}{\mu}$ ($\mu = 4/3$), so

$$\beta_w = \frac{\lambda_w D}{d} = \frac{\beta}{\mu} = \frac{3}{4}\beta$$

The fringe width decreases (to about 3/4 of its value in air).

i) Shifted from air into water: The wavelength of light in water is $\lambda_w = \lambda/\mu$, where $\mu \approx 4/3$. Since $\beta = \dfrac{\lambda D}{d}$, the fringe width decreases to $\beta/\mu \approx \tfrac{3}{4}\beta$. The fringes come closer together.

ii) Slit width increased: The fringe width $\beta = \lambda D/d$ depends on the separation $d$ of the slits, not on the width of each slit, so the fringe width does not change. However, the fringes become brighter but less distinct (the dark fringes are no longer fully dark), and if the slits are made very wide the pattern disappears.

Interference of light: When two coherent light waves superpose, the resultant intensity is redistributed, being maximum at some points and minimum at others. This gives alternate bright and dark fringes and is called interference of light.

Resultant intensity

For two coherent waves of intensities $I_1$ and $I_2$ with phase difference $\phi$:

$$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$

Here $I_1 = I_2 = I$:

$$I_R = 2I + 2I\cos\phi = 2I(1 + \cos\phi) = 4I\cos^2\frac{\phi}{2}$$

i) Constructive interference: $\phi = 0, 2\pi, 4\pi, \dots$, so $\cos\phi = 1$:

$$I_{\max} = 2I(1 + 1) = 4I$$

ii) Destructive interference: $\phi = \pi, 3\pi, \dots$, so $\cos\phi = -1$:

$$I_{\min} = 2I(1 - 1) = 0$$

Resultant intensity is $4I$ at a bright point and $0$ at a dark point. Energy is not lost; it is only redistributed (average intensity $= 2I$).

The colours are due to interference of light in thin films.

When white light falls on a soap film, part of it is reflected from the upper surface and part is refracted, reflected from the lower surface and comes out again. These two reflected rays come from the same incident ray, so they are coherent and interfere.

      incident   ray 1  ray 2
           \      /     /
  __________\____/_____/______  upper surface
             \  /\    /          film (μ, thickness t)
  ____________\/__\__/________  lower surface

The path difference between them is $2\mu t\cos r$ (plus $\lambda/2$ due to reflection at the denser surface). Constructive interference occurs for

$$2\mu t\cos r = \left(n + \tfrac{1}{2}\right)\lambda$$

This condition depends on the wavelength. At a given point of the film (given $t$ and $r$), only certain wavelengths (colours) are reinforced while others are weakened. Because the thickness of the film varies from place to place and the viewing angle changes, different colours are seen at different places, giving the beautiful colours of soap bubbles.

(i) Interference of light

Interference is the redistribution of light energy in space when two coherent waves superpose, giving alternate maxima (bright fringes) and minima (dark fringes) of intensity.

(ii) Fringe width

Let two coherent slits $S_1$, $S_2$ be a distance $d$ apart, and the screen be a distance $D$ away ($D \gg d$). For a point $P$ at height $y$ from the central point, the path difference is

$$\Delta = S_2P - S_1P \approx \frac{yd}{D}$$

For the $n$th bright fringe, $\Delta = n\lambda$, so $y_n = \dfrac{n\lambda D}{d}$. The fringe width is the distance between two consecutive bright (or dark) fringes:

$$\beta = y_{n+1} - y_n = \frac{\lambda D}{d}$$

(iii) Apparatus immersed in a liquid

In a liquid of refractive index $\mu$ the wavelength becomes $\lambda' = \lambda/\mu$, with $D$ and $d$ unchanged, so

$$\beta' = \frac{\lambda' D}{d} = \frac{\beta}{\mu}$$

The fringe width decreases by the factor $\mu$ (fringes become closer).

Given: $\lambda = 546$ nm $= 5.46\times10^{-7}$ m, $d = 0.50$ mm $= 5.0\times10^{-4}$ m, $D = 0.80$ m.

Fringe width:

$$\beta = \frac{\lambda D}{d} = \frac{5.46\times10^{-7} \times 0.80}{5.0\times10^{-4}} = 8.74\times10^{-4} \text{ m} = 0.874 \text{ mm}$$

Width of ten fringes:

$$10\beta = 10 \times 0.874 \text{ mm} = 8.74 \text{ mm}$$

The ten fringes cover about 8.7 mm.