Unit 06: Optics • Chapter 10 (Wave Optics)
Interference of Light & Young's Double Slit Experiment
Coherent sources, analytical superposition, path difference ($\Delta x = yd/D$), fringe width derivation ($\beta = \lambda D/d$), and cosine-squared intensity profiles.
Socho do speakers ko jo bilkul ek hi gaana, ek hi frequency aur zero time delay ke sath play kar rahe hain. Kamre ke kuch points par dono speakers ke sound crests ek sath jud kar double volume (loud sound) banayenge, jabki kuch points par pehle speaker ka crest doosre speaker ke trough ko 100% cancel karke sannata (dead silence) paida kar dega!
1801 se pehle log sochte the ki "Light + Light hamesha zyada Light hi banayegi." Thomas Young ne YDSE experiment se dikha diya ki Light + Light milkar Darkness bhi bana sakti hai! Energy destroy nahi hoti, balki dark zones se redistribute hokar bright zones mein concentrate ho jaati hai ($0 + 4I_0 = 2 \times 2I_0$). Yahi energy redistribution Interference kehlata hai!
1. Coherence & Superposition Principle
Two sources are coherent if they maintain a constant phase difference:
If $I_1 = I_2 = I_0$: $I = 4I_0 \cos^2(\phi/2)$. Two independent light sources can NEVER be coherent because atomic emission involves random phase jumps every $10^{-8}\text{ s}$.
2. Conditions for Maxima & Minima
Path difference $\Delta x = S_2 P - S_1 P$ and phase difference $\phi = \frac{2\pi}{\lambda}\Delta x$:
Where $n = 0, 1, 2, 3\dots$ ($n=0$ corresponds to Central Bright Fringe where path difference is strictly zero).
Analytical Derivation of Path Difference ($\Delta x = \frac{yd}{D}$) and Fringe Width ($\beta = \frac{\lambda D}{d}$)
Fig 10.2A: Standard Young's Double Slit Geometry showing path difference $S_2 P - S_1 P = S_2 N$.
1. Consider two narrow coherent slit sources $S_1$ and $S_2$ separated by small distance $d$. Let $D$ be the distance of the screen from the slit plane ($D \gg d$).
2. Let $P$ be an arbitrary point on the screen at vertical distance $y$ from central symmetry point $O$.
3. Drop perpendiculars from $S_1$ and $S_2$ on the screen, meeting at points $M$ and $N$ on both sides of $O$. Then $PM = y - \frac{d}{2}$ and $PN = y + \frac{d}{2}$.
4. Applying the Pythagorean theorem to right-angled triangles $\triangle S_1 M P$ and $\triangle S_2 N P$:
5. Subtracting the two equations:
6. Since $d \ll D$ and $y \ll D$, we can approximate $S_2 P \approx S_1 P \approx D \implies (S_2 P + S_1 P) \approx 2D$:
7. Position of $n$-th Bright Fringe (Constructive: $\Delta x = n\lambda$):
8. Fringe Width ($\beta$): The distance between two consecutive bright fringes:
Similarly, for dark fringes ($\Delta x = (2n-1)\frac{\lambda}{2} \implies y_n' = (2n-1)\frac{\lambda D}{2d}$), the dark fringe width is also $\beta' = y_n' - y_{n-1}' = \frac{\lambda D}{d}$. Hence, all bright and dark fringes in YDSE have identical, uniform width!
Mathematical Derivation of Intensity Distribution & Proof of Energy Conservation
1. Let two interfering light waves of equal amplitude $a$ and phase difference $\phi$ be represented by:
2. By the principle of superposition, resultant wave $y = y_1 + y_2 = a[\sin\omega t + \sin(\omega t + \phi)]$. Using $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:
3. Resultant amplitude is $R = 2a\cos(\phi/2)$. Since light intensity is proportional to the square of amplitude ($I \propto R^2$):
4. Proof of Law of Conservation of Energy:
• At Bright Fringes (Maxima): $\cos^2(\phi/2) = 1 \implies I_{\max} = 4I_0$.
• At Dark Fringes (Minima): $\cos^2(\phi/2) = 0 \implies I_{\min} = 0$.
• Average Intensity over a complete cycle: Since the average value of $\cos^2(\phi/2)$ is $\frac{1}{2}$:
Thermodynamic Truth: The average intensity is exactly equal to the sum of the individual intensities of the two slits ($2I_0$). Energy is not created at maxima nor destroyed at minima; it is simply redistributed from regions of destructive interference to constructive interference.
Trap 1: YDSE Apparatus Immersed in Water ($\mu$)
*Board Question:* What happens to the fringe width if the entire YDSE apparatus is immersed in water ($\mu = 4/3$)?
• Speed of light and wavelength decrease in water: $\lambda' = \frac{\lambda}{\mu}$.
• Slit separation $d$ and screen distance $D$ remain unchanged.
$$\beta' = \frac{\lambda' D}{d} = \frac{(\lambda/\mu) D}{d} = \mathbf{\frac{\beta}{\mu}}$$
Fringe width shrinks by a factor of $\mu$! (For water, fringes become $\frac{3}{4}\text{th}$ as wide, cramping closer together).
Trap 2: White Light Illumination in YDSE
*Exam Trap:* What pattern is observed on the screen if monochromatic light is replaced by white light?
1. At the center ($y = 0$), path difference is zero for ALL wavelengths ($\Delta x = 0$). Hence, all colors superpose constructively to form a CENTRAL WHITE FRINGE!
2. Because $\beta \propto \lambda$ and $\lambda_{\text{violet}} < \lambda_{\text{red}}$, the closest colored fringe on either side of central white is VIOLET, and the farthest visible fringe is RED.
3. After a few colored fringes, high orders overlap, producing uniform general illumination.
Trap 3: Introduction of a Thin Mica Sheet ($\mu, t$)
When a transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one slit:
• Additional optical path introduced $= (\mu - 1)t$.
• The entire fringe pattern shifts towards the side containing the sheet by distance:
$$y_0 = \frac{D}{d}(\mu - 1)t = \frac{\beta}{\lambda}(\mu - 1)t$$
• Crucial: Fringe width $\beta$ remains COMPLETELY UNCHANGED! Only the pattern shifts.
Trap 4: Ratio of Slit Widths to Fringe Contrast
Slit width $w \propto \text{Amplitude } a^2 \propto \text{Intensity } I$.
If ratio of slit widths is $w_1/w_2 = r$:
$$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \mathbf{\left( \frac{\sqrt{r} + 1}{\sqrt{r} - 1} \right)^2}$$
If one slit is completely closed ($I_2 = 0$), interference fringes vanish; only single-slit diffraction illumination remains!
Problem: In a Young’s double slit experiment, the slits are separated by $0.28\text{ mm}$ and the screen is placed $1.4\text{ m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\text{ cm}$. Determine the wavelength of light used in the experiment.
Solution:
1. Given: Slit separation $d = 0.28\text{ mm} = 0.28 \times 10^{-3}\text{ m}$, Screen distance $D = 1.4\text{ m}$.
2. Distance of 4th bright fringe ($n = 4$): $y_4 = 1.2\text{ cm} = 1.2 \times 10^{-2}\text{ m}$.
3. Solving for wavelength $\lambda$:
Problem: Two coherent light sources having intensity ratio $\beta_I = 81:1$ produce interference fringes. Calculate the ratio of maximum to minimum intensity in the interference pattern. Also, find the ratio of intensities at a point where path difference is $\lambda/4$ to that where path difference is $\lambda/3$.
Solution:
1. Given: $\frac{I_1}{I_2} = \frac{81}{1} \implies \frac{a_1}{a_2} = \sqrt{\frac{81}{1}} = \frac{9}{1}$.
2. General intensity formula for equal amplitude approximation ($I = 4I_0 \cos^2(\phi/2)$):
• For path difference $\Delta x_1 = \lambda/4 \implies \phi_1 = \frac{2\pi}{\lambda}\left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$ radians ($90^\circ$):
• For path difference $\Delta x_2 = \lambda/3 \implies \phi_2 = \frac{2\pi}{\lambda}\left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$ radians ($120^\circ$):
Path Difference
$\Delta x = \frac{yd}{D}$
Fringe Width
$\beta = \frac{\lambda D}{d}$ (Uniform)
Intensity Profile
$I = 4I_0 \cos^2(\phi/2)$
In Liquid ($\mu$)
$\beta' = \frac{\beta}{\mu}$ (Shrinks)