The Diffraction Grating: JEE's Favorite Tool for Spectral Analysis

Why Diffraction Grating is Crucial for JEE

Diffraction grating is one of the most frequently tested topics in JEE Physics, appearing in both theory and numerical problems. Understanding it helps in:

Wave optics numerical problems (JEE Main)
Conceptual understanding of interference patterns
Spectral analysis applications (JEE Advanced)
Resolving power calculations in modern physics

Concept 1 Fundamental

The Grating Equation: d sinθ = nλ

[Diagram: Diffraction grating showing incident light, slits spacing d, and diffraction angle θ]

📐 Derivation of d sinθ = nλ:

Step 1: Consider two adjacent slits in the grating

Step 2: Path difference between waves from adjacent slits:

$ \Delta = d \sin\theta $

Step 3: For constructive interference (bright fringe):

$ \Delta = n\lambda $ where $n = 0, \pm1, \pm2, \ldots$

Step 4: Combine both conditions:

$$ d \sin\theta = n\lambda $$

Where:
• $d$ = grating spacing
• $\theta$ = diffraction angle
• $n$ = order of spectrum
• $\lambda$ = wavelength of light

🎯 JEE Application Example:

Problem: A diffraction grating has 5000 lines/cm. Find the angle for first order maximum for light of wavelength 600 nm.

Solution:

Grating spacing $d = \frac{1}{5000}$ cm $= 2 \times 10^{-6}$ m

Using $d \sin\theta = n\lambda$:

$\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 600 \times 10^{-9}}{2 \times 10^{-6}} = 0.3$

$\theta = \sin^{-1}(0.3) \approx 17.46^\circ$

Concept 2 Important

Formation of Spectra

🌈 Understanding Spectral Orders:

White Light Spectrum:

• For white light, each wavelength forms its own set of maxima

• Central maximum (n=0) is white - all wavelengths overlap

• First order (n=±1) shows complete spectrum from violet to red

• Higher orders show more spreading but overlapping occurs

Angular Dispersion:

Differentiation of grating equation:

$ d \cos\theta d\theta = n d\lambda $

Angular dispersion $ = \frac{d\theta}{d\lambda} = \frac{n}{d \cos\theta} $

🎯 JEE Application Example:

Problem: In a grating spectrum, which color deviates more and why?

Solution:

From $d \sin\theta = n\lambda$, for fixed n and d:

$\sin\theta \propto \lambda$

Since red light has higher wavelength ($\lambda_R > \lambda_V$),

$\theta_R > \theta_V$ ⇒ Red light deviates more

Concept 3 Advanced

Resolving Power of Grating

🔍 Rayleigh Criterion and Resolving Power:

Rayleigh Criterion:

Two spectral lines are just resolved when principal maximum of one falls on first minimum of other

Derivation of Resolving Power:

Step 1: Angular width of principal maximum:

$ \Delta\theta = \frac{\lambda}{Nd\cos\theta} $

Step 2: From grating equation differentiation:

$ d\cos\theta \Delta\theta = n \Delta\lambda $

Step 3: Equating both expressions:

$ \frac{\lambda}{Nd\cos\theta} = \frac{n \Delta\lambda}{d\cos\theta} $

Step 4: Resolving power:

$$ R = \frac{\lambda}{\Delta\lambda} = nN $$

Where N = total number of rulings

🎯 JEE Application Example:

Problem: A grating has 10000 lines and is used in second order. Calculate minimum wavelength difference it can resolve for λ = 500 nm.

Solution:

Resolving power $R = nN = 2 \times 10000 = 20000$

$ R = \frac{\lambda}{\Delta\lambda} $

$ \Delta\lambda = \frac{\lambda}{R} = \frac{500}{20000} = 0.025$ nm

Concept 4 Advanced

Missing Orders in Grating Spectrum

🚫 Condition for Missing Orders:

Single Slit Diffraction Effect:

• Grating pattern = Interference pattern × Single slit diffraction pattern

• When interference maximum coincides with diffraction minimum, that order is missing

Mathematical Condition:

Interference maximum: $d \sin\theta = n\lambda$

Diffraction minimum: $a \sin\theta = m\lambda$

Combining: $ \frac{d}{a} = \frac{n}{m} $

Where a = slit width, d = grating spacing

Missing orders occur when n = m × (d/a)

🎯 JEE Application Example:

Problem: A grating has slit width a = 1 μm and spacing d = 3 μm. Which orders are missing?

Solution:

$ \frac{d}{a} = \frac{3}{1} = 3 $

Missing orders occur when n = 3, 6, 9, ...

So 3rd, 6th, 9th... orders will be missing

🚀 Problem-Solving Strategies

Key Formulas to Remember:

Grating equation: $d \sin\theta = n\lambda$
Angular dispersion: $\frac{d\theta}{d\lambda} = \frac{n}{d\cos\theta}$
Resolving power: $R = nN$
Missing orders: $\frac{d}{a} = \frac{n}{m}$

JEE Exam Tips:

Always check units (nm vs m)
Maximum order: $n_{max} = \frac{d}{\lambda}$
For white light, central max is white
Higher orders have better resolution

Advanced Applications Available

Includes overlapping orders, grating with oblique incidence, and JEE Advanced level problems

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. A grating has 6000 lines/cm. Find wavelength if 2nd order maximum is at 45°.

2. Calculate resolving power of grating with 15000 lines used in 3rd order.

3. If d/a = 2.5, which orders are missing in the spectrum?

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