Diffraction Grating Numericals: How to Find Wavelength, Order, and More
Master the most common JEE question types with step-by-step problem-solving approaches and shortcuts.
Why Master Diffraction Grating Numericals?
Diffraction grating problems appear in 85% of JEE Main papers and are essential for JEE Advanced. These numericals test your understanding of:
- Grating formula application under different conditions
- Wave optics concepts and interference principles
- Mathematical manipulation of trigonometric equations
- Practical understanding of optical instruments
The Fundamental Grating Formula
Where:
$a + b$ = Grating element (width of slit + spacing)
$\theta$ = Diffraction angle
$n$ = Order of spectrum
$\lambda$ = Wavelength of light
Key Points:
• $n = 0, 1, 2, 3, ...$ (Order number)
• Maximum order: $n_{max} = \frac{a+b}{\lambda}$
• $\sin\theta \leq 1$ (Physical constraint)
Finding Wavelength of Light
Given grating parameters and diffraction angle, find the wavelength.
📝 Problem Statement:
A diffraction grating has 5000 lines per cm. The second order spectrum is observed at an angle of 30°. Find the wavelength of light used.
🧮 Step-by-Step Solution:
Step 1: Find grating element $(a + b)$
Number of lines = 5000 per cm = 500,000 per meter
$(a + b) = \frac{1}{500000} = 2 \times 10^{-6}$ m
Step 2: Apply grating formula
$(a + b)\sin\theta = n\lambda$
$2 \times 10^{-6} \times \sin 30° = 2 \times \lambda$
Step 3: Solve for wavelength
$2 \times 10^{-6} \times 0.5 = 2\lambda$
$1 \times 10^{-6} = 2\lambda$
$\lambda = 5 \times 10^{-7}$ m = 500 nm
Finding Maximum Possible Order
Determine the highest order spectrum visible for given grating and wavelength.
📝 Problem Statement:
A diffraction grating has 6000 lines per cm. Find the maximum order for light of wavelength 600 nm.
🧮 Step-by-Step Solution:
Step 1: Find grating element
$(a + b) = \frac{1}{6000 \times 100} = 1.667 \times 10^{-6}$ m
Step 2: Use maximum order condition
$(a + b)\sin\theta = n\lambda$ and $\sin\theta \leq 1$
Maximum when $\sin\theta = 1$
$n_{max} = \frac{a + b}{\lambda}$
Step 3: Calculate maximum order
$n_{max} = \frac{1.667 \times 10^{-6}}{600 \times 10^{-9}} = 2.778$
Since order must be integer: $n_{max} = 2$
Finding Diffraction Angle
Calculate the angle at which a particular order is observed.
📝 Problem Statement:
A grating with 4000 lines/cm is used with light of wavelength 500 nm. Find the angle for first order maximum.
🧮 Step-by-Step Solution:
Step 1: Find grating element
$(a + b) = \frac{1}{4000 \times 100} = 2.5 \times 10^{-6}$ m
Step 2: Apply grating formula
$(a + b)\sin\theta = n\lambda$
$2.5 \times 10^{-6} \times \sin\theta = 1 \times 500 \times 10^{-9}$
Step 3: Solve for angle
$\sin\theta = \frac{500 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.2$
$\theta = \sin^{-1}(0.2) = 11.54°$
🚀 Problem-Solving Shortcuts
Unit Conversion Tips:
- 1 cm = 0.01 m (for grating element)
- 1 nm = $10^{-9}$ m (for wavelength)
- Lines/cm to lines/m: multiply by 100
- Grating element = 1/(lines per meter)
Common Mistakes to Avoid:
- Forgetting $\sin\theta \leq 1$ constraint
- Unit inconsistency (nm vs m)
- Taking fractional order values
- Missing the integer condition for n
💡 Important Physical Insights
• Angular Dispersion: Rate of change of angle with wavelength $\frac{d\theta}{d\lambda} = \frac{n}{(a+b)\cos\theta}$
• Resolving Power: $R = \frac{\lambda}{d\lambda} = nN$ (where N = total lines)
• Overlapping Orders: $n_1\lambda_1 = n_2\lambda_2$ for overlapping spectra
• Missing Orders: Occur when diffraction minimum coincides with interference maximum
More Problem Types Available
Includes overlapping orders, resolving power, angular dispersion, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. A grating has 8000 lines/cm. Find wavelength if 3rd order is at 45°.
2. For λ=550 nm and 10000 lines/cm, find maximum visible order.
3. Calculate angular separation between 400 nm and 700 nm in first order.
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