Unit 06: Optics • Chapter 10 (Wave Optics)
Diffraction of Light: Single Slit & Central Maximum
Bending around obstacles, wave division technique, conditions for minima & secondary maxima, central peak angular width, and ray optics validity (Fresnel distance).
Socho ek deewar ke peeche khada vyakti hume dikhayi nahi deta, par uski aawaz saaf sunayi deti hai. Aisa isliye kyunki sound waves ki wavelength ($\sim 1\text{ m}$) darwaze aur deewar ke size ke barabar hoti hai, jisse sound aasaani se corners ke chaaron taraf bend ho jaati hai. Light waves ke sath aamtaur par aisa nahi hota kyunki light ki wavelength bohot microscopic hoti hai ($\sim 10^{-7}\text{ m}$). Lekin jab light ko ek baal jitni patli slit ($a \approx \lambda$) se guzara jata hai, toh light seedhe chalna chhodkar geometric shadow ke andar mud jaati hai! Yahi bending Diffraction hai. YDSE mein do alag-alag slits ke wavelets aapas mein lad rahe the; yahan ek hi slit ke andar baithe hazaron secondary wavelets aapas mein interfere karte hain!
1. Minima & Secondary Maxima Positions
Slit width $a$, diffraction angle $\theta$, and wavelength $\lambda$:
Notice the inversion: $a\sin\theta = n\lambda$ gives dark minima in single-slit diffraction, whereas it gave bright maxima in YDSE!
2. Central Maximum Width & Fresnel Distance
Spread of the intense central peak bounded by first minima on either side:
Ray optics remains valid for travel distances $D < Z_F$. When $D \gg Z_F$, diffraction spread dominates over rectilinear propagation.
Deduction of Minima ($a\sin\theta = n\lambda$), Maxima, and Wavefront Division
Fig 10.3A: Fraunhofer single slit diffraction geometry showing slit aperture $AB = a$ and path difference $BN = a\sin\theta$.
1. Consider a plane wavefront incident normally on a slit $AB$ of width $a$. In accordance with Huygens' principle, every continuous point inside the slit becomes a source of secondary wavelets.
2. At the central point $O$ on the screen, wavelets from symmetrical elements across the entire slit travel identical optical path lengths ($\theta = 0 \implies \Delta x = 0$). All wavelets arrive in-phase, forming an intensely bright Central Maximum.
3. Consider wavelets diffracted at angle $\theta$ converging at point $P$. The total path difference between extreme rays from $A$ and $B$ is:
4. Condition for Minima (Destructive Interference):
• Let $a\sin\theta = \lambda$. Divide the slit $AB$ into two equal halves $AM$ and $MB$, each of width $a/2$.
• For every point in the upper half $AM$, there exists a corresponding point in the lower half $MB$ separated by distance $a/2$, whose path difference is:
• The wavelets from these pairs cancel each other out pairwise completely, producing zero net resultant intensity! Hence, the general condition for $n$-th minimum is:
5. Condition for Secondary Maxima:
• Let $a\sin\theta = \frac{3\lambda}{2}$. Divide the slit into three equal parts, each of width $a/3$.
• Wavelets from the first two parts cancel each other in pairs ($\Delta x = \lambda/2$). However, the wavelets from the remaining third part survive and send light to the screen, forming the first secondary maximum (with intensity only $\approx \frac{1}{22}$ of the central peak).
• General condition for $n$-th secondary maximum:
Width of Central Maximum and Validity of Ray Optics (Fresnel Distance $Z_F$)
1. The Central Maximum extends on either side of the center $O$ up to the first minima ($n = 1$):
2. The total Angular Width of the central maximum is the separation between first minima on both sides:
3. If $D$ is the distance to the screen, the linear half-width is $y_1 = D \theta_1 = \frac{\lambda D}{a}$. Hence, the total Linear Width of the central maximum:
4. Fresnel Distance ($Z_F$): As a beam of width $a$ propagates over distance $z$, diffraction causes it to spread by an amount $y \approx z \theta = \frac{z\lambda}{a}$. The distance at which this diffraction spread equals the size of the aperture itself ($y = a$) is the Fresnel Distance:
Boundary Meaning: For travel distances $z \ll Z_F$, diffraction spreading is completely negligible and rectilinear ray optics is rigorously valid. For $z \gg Z_F$, wave optics must be applied.
Master Comparison: Interference (YDSE) vs Diffraction (Single Slit)
| Feature | Interference (YDSE) | Diffraction (Single Slit) |
|---|---|---|
| Origin of Waves | Superposition of two waves originating from two discrete slits ($S_1, S_2$). | Superposition of continuous secondary wavelets from different parts of the same single slit. |
| Fringe Width | All fringes are of EQUAL width ($\beta = \frac{\lambda D}{d}$). | Central peak is TWICE as wide ($2\beta_0 = \frac{2\lambda D}{a}$); higher fringes become narrower. |
| Intensity of Maxima | All bright fringes have equal maximum intensity ($I_{\max} = 4I_0$). | Central maximum is extremely bright; secondary maxima drop drastically ($I_0 : \frac{I_0}{22} : \frac{I_0}{61}$). |
| Minima Darkness | Minima are perfectly dark ($I_{\min} = 0$, high contrast). | Minima are rarely completely dark due to incomplete wavelet cancellation. |
Trap 1: What Happens When Slit Width $a$ is Made Very Large?
*Board Question:* If the single slit width $a$ is widened to $1\text{ cm}$, what pattern is observed?
• Angular width of central peak is $2\theta = \frac{2\lambda}{a}$.
• As $a$ increases, $2\theta \to 0$. The diffraction spread completely vanishes!
• Diffraction pattern disappears completely, and light casts a sharp, clear geometric shadow, behaving strictly as a classical ray!
Trap 2: The $a\sin\theta = n\lambda$ Condition Reversal Trap
*Exam Confusion Alert:*
• In Interference (YDSE): $\Delta x = n\lambda$ is the condition for BRIGHT MAXIMA.
• In Diffraction (Single Slit): $a\sin\theta = n\lambda$ is the condition for DARK MINIMA!
Never mix these up. In diffraction, $n\lambda$ allows dividing the slit into $2n$ equal zones that cancel in pairs.
Trap 3: Central Maximum With White Light Illumination
When white light illuminates a single slit:
1. The center of the screen ($\theta = 0$) has zero path difference for all colors; hence it forms a CENTRAL WHITE MAXIMUM.
2. Because $\beta_0 \propto \lambda$, the red boundary spreads wider than violet.
3. The inner edge of the first diffraction fringe is VIOLET, and the outer edge is RED.
Trap 4: Missing Orders in Combined Interference-Diffraction
In actual double slit experiments, diffraction envelope modulates interference fringes. If $n$-th interference bright fringe coincides with $m$-th diffraction minimum: $$\frac{n\lambda D}{d} = \frac{m\lambda D}{a} \implies \mathbf{n = m\left(\frac{d}{a}\right)}$$ These specific interference fringes will be MISSING (absent) from the pattern!
Problem: A parallel beam of light of wavelength $500\text{ nm}$ falls normally on a single narrow slit of width $0.20\text{ mm}$. The resulting diffraction pattern is observed on a screen placed $2.0\text{ m}$ away. Calculate (i) the angular width of the central maximum, (ii) the linear width of the central maximum, and (iii) the distance of the second dark band from the center of the screen.
Solution: $\lambda = 500\text{ nm} = 5.0 \times 10^{-7}\text{ m}$, slit width $a = 0.20\text{ mm} = 2.0 \times 10^{-4}\text{ m}$, $D = 2.0\text{ m}$.
1. Angular width of central maximum ($2\theta_1$):
2. Linear width of central maximum ($2\beta_0$):
3. Distance of second dark band ($n = 2$ minimum):
Problem: For what distance is ray optics a good approximation when the aperture is $3\text{ mm}$ wide and the wavelength is $500\text{ nm}$? If the screen is placed at $D = 30\text{ m}$, calculate the spread of the beam due to diffraction and compare it with the size of the aperture.
Solution: Aperture $a = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}$, $\lambda = 500\text{ nm} = 5.0 \times 10^{-7}\text{ m}$.
1. Fresnel distance ($Z_F$):
Ray optics is a good approximation for distances up to $18.0\text{ metres}$.
2. At $D = 30\text{ m}$ (which is $> Z_F$), the linear spread due to diffraction is:
Notice that diffraction spread ($\Delta y = 5\text{ mm}$) exceeds the slit size ($a = 3\text{ mm}$). Total spot size $= a + 2\Delta y = 3 + 10 = 13\text{ mm}$.
Minima Condition
$a\sin\theta = n\lambda$
Secondary Maxima
$a\sin\theta = (2n+1)\frac{\lambda}{2}$
Central Width
$2\beta_0 = \frac{2\lambda D}{a}$
Fresnel Distance
$Z_F = \frac{a^2}{\lambda}$