Narrow Slit
After a narrow slit, light spreads into the geometrical-shadow region.
Wave Optics · Chapter 04
Diffraction of Light
Understand single-slit spreading, minima, secondary maxima, central width and resolution through exam-ready derivations and textbook-style diagrams.
CBSENEETJEE MainJEE AdvancedIB PhysicsIGCSEA-Level
DiffractionFresnel and Fraunhofer DiffractionSingle Slit DiffractionCentral MaximumWidth of Central MaximumSecondary Maxima and MinimaResolving PowerRayleigh CriterionApplicationsNumericalsPYQs
01
Diffraction of Light
Diffraction is the bending and spreading of light when it passes through a narrow aperture or around an obstacle. It becomes clearly observable when the aperture size is comparable with the wavelength.
a ≈ λ → significant diffractionSound Waves
Sound wavelengths are comparable with doors and walls, so sound bends readily.
Visible Light
Visible wavelengths are tiny compared with everyday openings, so spreading is normally small.
Diffraction shows the limitation of straight-line ray optics and directly supports the wave nature of light.
02
Huygens Principle and Diffraction
Every point within the open slit emits a secondary wavelet. At a distant point, these wavelets arrive with different phases; their interference produces the observed maxima and minima.
03
Fresnel and Fraunhofer Diffraction
| Feature | Fresnel Diffraction | Fraunhofer Diffraction |
|---|---|---|
| Source and screen | Finite distances | Effectively infinite distances or focal planes |
| Wavefront | Spherical/curved | Plane incident wavefront |
| Lenses | Not essential | Usually used for collimation and focusing |
| Pattern | Changes with distance | Stable angular pattern |
| Application | Near-field edge patterns | Single slit, gratings, spectroscopy |
04
Single Slit Diffraction
Consider a slit AB of width a illuminated by a plane monochromatic wave.
For rays leaving A and B at angle θ, the extreme path difference is Δ = a sinθ.
The resultant amplitude is the vector sum of contributions from all elemental strips.
At θ = 0 all contributions are in phase, giving the principal maximum.
05
Position of Minima
First minimum: a sinθ = λ. Divide the slit into two halves; corresponding points differ by λ/2 and cancel pairwise.
Second minimum: a sinθ = 2λ. Divide the slit into four equal parts; adjacent paired contributions again differ by λ/2.
In general, complete cancellation occurs for a sinθ = nλ, n = 1,2,3,...
06
Position of Secondary Maxima
For the first secondary maximum, a sinθ ≈ 3λ/2.
Divide the aperture into three parts. Two parts cancel approximately and one part remains.
Similarly the next weak maximum occurs near a sinθ ≈ 5λ/2.
The simple exam-level approximate condition is a sinθ = (2n+1)λ/2.
This odd-half-wave condition is an approximation. Exact secondary maxima satisfy tanα = α, where α = πa sinθ/λ.
07
Central Maximum
The first minima lie at a sinθ = ±λ.
For small angles, θ ≈ λ/a.
The central maximum extends from −θ to +θ, so angular width = 2θ = 2λ/a.
At screen distance D, half-width y₁ = Dλ/a.
Therefore linear width β₀ = 2y₁ = 2Dλ/a.
Angular width = 2λ/aLinear width β₀ = 2Dλ/a
08
Intensity Distribution Curve
- The central maximum has the greatest intensity because all wavelets are in phase at θ = 0.
- Secondary maxima rapidly decrease because cancellation becomes increasingly effective.
- Minima occur at ±λ/a, ±2λ/a, ±3λ/a, ... for small angles.
09
Width of Central Maximum
β₀ = 2Dλ/a
| Change | Effect |
|---|---|
| Increase λ | Central maximum widens |
| Increase D | Linear width increases |
| Decrease a | Central maximum widens |
Example: If D doubles while λ and a remain fixed, β₀ doubles.
10
Resolving Power
Telescope
Minimum angular separation θmin = 1.22λ/d for a circular aperture. Larger objective diameter improves resolution.
Microscope
Limit of resolution is approximately 0.61λ/(μ sinα). Shorter wavelength and larger numerical aperture improve detail.
Human Eye
The finite pupil causes diffraction; a typical angular resolution is of order one arcminute under good conditions.
Resolving power is the ability to distinguish nearby objects or spectral features as separate.
11
Rayleigh Criterion
Two sources are just resolved when the central maximum of either diffraction image falls at the first minimum of the other. Smaller separation gives strongly overlapping images; larger separation gives clearly distinct peaks.
12
Applications of Diffraction
CD/DVD colors
Closely spaced tracks act like a diffraction grating and separate colors.
Optical instruments
Aperture diffraction fixes the ultimate image sharpness.
Spectroscopy
Gratings separate wavelengths for precise spectral analysis.
X-ray diffraction
Crystal planes diffract X-rays and reveal atomic structure.
Astronomy
Diffraction limits angular resolution of stars and planets.
Microscopes and telescopes
Aperture size and wavelength set the smallest resolvable detail.
13
Important Formula Sheet
Δ = a sinθa sinθ = nλa sinθ ≈ (2n+1)λ/2Angular width = 2λ/aLinear width β₀ = 2Dλ/aTelescope: θmin = 1.22λ/d
14
30 Solved Numericals
NEET
1. First minimum
For slit width a = 0.2 mm and λ = 400 nm, find the small angle of first minimum.
Answer: 2.000e-3 rad
Solution: a sinθ = λ. For small θ, θ = λ/a = 2.000e-3 rad.
NEET
2. Central width
Find β₀ for D = 1.5 m, λ = 550 nm and a = 0.5 mm.
Answer: 3.300 mm
Solution: β₀ = 2Dλ/a = 3.300 mm.
NEET
3. Nth minimum
Write the path condition for the 3rd minimum.
Answer: a sinθ = 3λ
Solution: The general minimum condition is a sinθ = nλ; substituting n = 3 gives the result.
NEET
4. Secondary maximum
State the approximate condition for secondary maximum number 1.
Answer: a sinθ = 3λ/2
Solution: Using the conceptual odd-half-wave condition a sinθ = (2n+1)λ/2 gives 3λ/2.
NEET
5. First minimum position
For D = 1.5 m, λ = 600 nm and a = 0.3 mm, find the distance of first minimum from center.
Answer: 3.000 mm
Solution: y₁ = Dλ/a = 3.000 mm; central width is twice this value.
NEET
6. Parameter change
If wavelength is multiplied by 1.25, what happens to central width?
Answer: It becomes 1.25 times
Solution: β₀ = 2Dλ/a, so central width changes in direct proportion to λ.
NEET
7. First minimum
For slit width a = 0.3 mm and λ = 400 nm, find the small angle of first minimum.
Answer: 1.333e-3 rad
Solution: a sinθ = λ. For small θ, θ = λ/a = 1.333e-3 rad.
NEET
8. Central width
Find β₀ for D = 2.5 m, λ = 650 nm and a = 0.5 mm.
Answer: 6.500 mm
Solution: β₀ = 2Dλ/a = 6.500 mm.
NEET
9. Nth minimum
Write the path condition for the 1st minimum.
Answer: a sinθ = 1λ
Solution: The general minimum condition is a sinθ = nλ; substituting n = 1 gives the result.
NEET
10. Secondary maximum
State the approximate condition for secondary maximum number 1.
Answer: a sinθ = 3λ/2
Solution: Using the conceptual odd-half-wave condition a sinθ = (2n+1)λ/2 gives 3λ/2.
JEE Main
11. First minimum position
For D = 1.5 m, λ = 600 nm and a = 0.3 mm, find the distance of first minimum from center.
Answer: 3.000 mm
Solution: y₁ = Dλ/a = 3.000 mm; central width is twice this value.
JEE Main
12. Parameter change
If wavelength is multiplied by 1.75, what happens to central width?
Answer: It becomes 1.75 times
Solution: β₀ = 2Dλ/a, so central width changes in direct proportion to λ.
JEE Main
13. First minimum
For slit width a = 0.4 mm and λ = 400 nm, find the small angle of first minimum.
Answer: 1.000e-3 rad
Solution: a sinθ = λ. For small θ, θ = λ/a = 1.000e-3 rad.
JEE Main
14. Central width
Find β₀ for D = 1.5 m, λ = 550 nm and a = 0.5 mm.
Answer: 3.300 mm
Solution: β₀ = 2Dλ/a = 3.300 mm.
JEE Main
15. Nth minimum
Write the path condition for the 3rd minimum.
Answer: a sinθ = 3λ
Solution: The general minimum condition is a sinθ = nλ; substituting n = 3 gives the result.
JEE Main
16. Secondary maximum
State the approximate condition for secondary maximum number 1.
Answer: a sinθ = 3λ/2
Solution: Using the conceptual odd-half-wave condition a sinθ = (2n+1)λ/2 gives 3λ/2.
JEE Main
17. First minimum position
For D = 1.5 m, λ = 600 nm and a = 0.3 mm, find the distance of first minimum from center.
Answer: 3.000 mm
Solution: y₁ = Dλ/a = 3.000 mm; central width is twice this value.
JEE Main
18. Parameter change
If wavelength is multiplied by 1.25, what happens to central width?
Answer: It becomes 1.25 times
Solution: β₀ = 2Dλ/a, so central width changes in direct proportion to λ.
JEE Main
19. First minimum
For slit width a = 0.5 mm and λ = 400 nm, find the small angle of first minimum.
Answer: 8.000e-4 rad
Solution: a sinθ = λ. For small θ, θ = λ/a = 8.000e-4 rad.
JEE Main
20. Central width
Find β₀ for D = 2.5 m, λ = 650 nm and a = 0.5 mm.
Answer: 6.500 mm
Solution: β₀ = 2Dλ/a = 6.500 mm.
JEE Main
21. Nth minimum
Write the path condition for the 1st minimum.
Answer: a sinθ = 1λ
Solution: The general minimum condition is a sinθ = nλ; substituting n = 1 gives the result.
JEE Main
22. Secondary maximum
State the approximate condition for secondary maximum number 1.
Answer: a sinθ = 3λ/2
Solution: Using the conceptual odd-half-wave condition a sinθ = (2n+1)λ/2 gives 3λ/2.
JEE Advanced
23. First minimum position
For D = 1.5 m, λ = 600 nm and a = 0.3 mm, find the distance of first minimum from center.
Answer: 3.000 mm
Solution: y₁ = Dλ/a = 3.000 mm; central width is twice this value.
JEE Advanced
24. Parameter change
If wavelength is multiplied by 1.75, what happens to central width?
Answer: It becomes 1.75 times
Solution: β₀ = 2Dλ/a, so central width changes in direct proportion to λ.
JEE Advanced
25. First minimum
For slit width a = 0.6 mm and λ = 400 nm, find the small angle of first minimum.
Answer: 6.667e-4 rad
Solution: a sinθ = λ. For small θ, θ = λ/a = 6.667e-4 rad.
JEE Advanced
26. Central width
Find β₀ for D = 1.5 m, λ = 550 nm and a = 0.5 mm.
Answer: 3.300 mm
Solution: β₀ = 2Dλ/a = 3.300 mm.
JEE Advanced
27. Nth minimum
Write the path condition for the 3rd minimum.
Answer: a sinθ = 3λ
Solution: The general minimum condition is a sinθ = nλ; substituting n = 3 gives the result.
JEE Advanced
28. Secondary maximum
State the approximate condition for secondary maximum number 1.
Answer: a sinθ = 3λ/2
Solution: Using the conceptual odd-half-wave condition a sinθ = (2n+1)λ/2 gives 3λ/2.
JEE Advanced
29. First minimum position
For D = 1.5 m, λ = 600 nm and a = 0.3 mm, find the distance of first minimum from center.
Answer: 3.000 mm
Solution: y₁ = Dλ/a = 3.000 mm; central width is twice this value.
JEE Advanced
30. Parameter change
If wavelength is multiplied by 1.25, what happens to central width?
Answer: It becomes 1.25 times
Solution: β₀ = 2Dλ/a, so central width changes in direct proportion to λ.
15
50 PYQ-Pattern Questions
Academic note: These are original CBSE, NEET, JEE, IB, IGCSE and A-Level exam-style questions, not verbatim copyrighted papers.
Q1For observable diffraction, choose the correct diffraction statement.
Answer: A. a ≈ λ
Explanation: The correct result is a ≈ λ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q2For path difference, choose the correct diffraction statement.
Answer: D. Δ = a sinθ
Explanation: The correct result is Δ = a sinθ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q3For minima, choose the correct diffraction statement.
Answer: C. a sinθ = nλ
Explanation: The correct result is a sinθ = nλ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q4For approximate secondary maxima, choose the correct diffraction statement.
Answer: B. a sinθ = (2n+1)λ/2
Explanation: The correct result is a sinθ = (2n+1)λ/2. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q5For angular central width, choose the correct diffraction statement.
Answer: A. 2λ/a
Explanation: The correct result is 2λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q6For linear central width, choose the correct diffraction statement.
Answer: D. 2Dλ/a
Explanation: The correct result is 2Dλ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q7For first minimum, choose the correct diffraction statement.
Answer: C. sinθ = λ/a
Explanation: The correct result is sinθ = λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q8For central maximum, choose the correct diffraction statement.
Answer: B. brightest and widest
Explanation: The correct result is brightest and widest. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q9For Fraunhofer source rays, choose the correct diffraction statement.
Answer: A. parallel
Explanation: The correct result is parallel. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q10For Rayleigh criterion, choose the correct diffraction statement.
Answer: D. central maximum coincides with first minimum
Explanation: The correct result is central maximum coincides with first minimum. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q11For observable diffraction, choose the correct diffraction statement.
Answer: C. a ≈ λ
Explanation: The correct result is a ≈ λ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q12For path difference, choose the correct diffraction statement.
Answer: B. Δ = a sinθ
Explanation: The correct result is Δ = a sinθ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q13For minima, choose the correct diffraction statement.
Answer: A. a sinθ = nλ
Explanation: The correct result is a sinθ = nλ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q14For approximate secondary maxima, choose the correct diffraction statement.
Answer: D. a sinθ = (2n+1)λ/2
Explanation: The correct result is a sinθ = (2n+1)λ/2. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q15For angular central width, choose the correct diffraction statement.
Answer: C. 2λ/a
Explanation: The correct result is 2λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q16For linear central width, choose the correct diffraction statement.
Answer: B. 2Dλ/a
Explanation: The correct result is 2Dλ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q17For first minimum, choose the correct diffraction statement.
Answer: A. sinθ = λ/a
Explanation: The correct result is sinθ = λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q18For central maximum, choose the correct diffraction statement.
Answer: D. brightest and widest
Explanation: The correct result is brightest and widest. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q19For Fraunhofer source rays, choose the correct diffraction statement.
Answer: C. parallel
Explanation: The correct result is parallel. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q20For Rayleigh criterion, choose the correct diffraction statement.
Answer: B. central maximum coincides with first minimum
Explanation: The correct result is central maximum coincides with first minimum. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q21For observable diffraction, choose the correct diffraction statement.
Answer: A. a ≈ λ
Explanation: The correct result is a ≈ λ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q22For path difference, choose the correct diffraction statement.
Answer: D. Δ = a sinθ
Explanation: The correct result is Δ = a sinθ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q23For minima, choose the correct diffraction statement.
Answer: C. a sinθ = nλ
Explanation: The correct result is a sinθ = nλ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q24For approximate secondary maxima, choose the correct diffraction statement.
Answer: B. a sinθ = (2n+1)λ/2
Explanation: The correct result is a sinθ = (2n+1)λ/2. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q25For angular central width, choose the correct diffraction statement.
Answer: A. 2λ/a
Explanation: The correct result is 2λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q26For linear central width, choose the correct diffraction statement.
Answer: D. 2Dλ/a
Explanation: The correct result is 2Dλ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q27For first minimum, choose the correct diffraction statement.
Answer: C. sinθ = λ/a
Explanation: The correct result is sinθ = λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q28For central maximum, choose the correct diffraction statement.
Answer: B. brightest and widest
Explanation: The correct result is brightest and widest. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q29For Fraunhofer source rays, choose the correct diffraction statement.
Answer: A. parallel
Explanation: The correct result is parallel. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q30For Rayleigh criterion, choose the correct diffraction statement.
Answer: D. central maximum coincides with first minimum
Explanation: The correct result is central maximum coincides with first minimum. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q31For observable diffraction, choose the correct diffraction statement.
Answer: C. a ≈ λ
Explanation: The correct result is a ≈ λ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q32For path difference, choose the correct diffraction statement.
Answer: B. Δ = a sinθ
Explanation: The correct result is Δ = a sinθ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q33For minima, choose the correct diffraction statement.
Answer: A. a sinθ = nλ
Explanation: The correct result is a sinθ = nλ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q34For approximate secondary maxima, choose the correct diffraction statement.
Answer: D. a sinθ = (2n+1)λ/2
Explanation: The correct result is a sinθ = (2n+1)λ/2. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q35For angular central width, choose the correct diffraction statement.
Answer: C. 2λ/a
Explanation: The correct result is 2λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q36For linear central width, choose the correct diffraction statement.
Answer: B. 2Dλ/a
Explanation: The correct result is 2Dλ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q37For first minimum, choose the correct diffraction statement.
Answer: A. sinθ = λ/a
Explanation: The correct result is sinθ = λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q38For central maximum, choose the correct diffraction statement.
Answer: D. brightest and widest
Explanation: The correct result is brightest and widest. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q39For Fraunhofer source rays, choose the correct diffraction statement.
Answer: C. parallel
Explanation: The correct result is parallel. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q40For Rayleigh criterion, choose the correct diffraction statement.
Answer: B. central maximum coincides with first minimum
Explanation: The correct result is central maximum coincides with first minimum. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q41For observable diffraction, choose the correct diffraction statement.
Answer: A. a ≈ λ
Explanation: The correct result is a ≈ λ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q42For path difference, choose the correct diffraction statement.
Answer: D. Δ = a sinθ
Explanation: The correct result is Δ = a sinθ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q43For minima, choose the correct diffraction statement.
Answer: C. a sinθ = nλ
Explanation: The correct result is a sinθ = nλ. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q44For approximate secondary maxima, choose the correct diffraction statement.
Answer: B. a sinθ = (2n+1)λ/2
Explanation: The correct result is a sinθ = (2n+1)λ/2. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q45For angular central width, choose the correct diffraction statement.
Answer: A. 2λ/a
Explanation: The correct result is 2λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q46For linear central width, choose the correct diffraction statement.
Answer: D. 2Dλ/a
Explanation: The correct result is 2Dλ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q47For first minimum, choose the correct diffraction statement.
Answer: C. sinθ = λ/a
Explanation: The correct result is sinθ = λ/a. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q48For central maximum, choose the correct diffraction statement.
Answer: B. brightest and widest
Explanation: The correct result is brightest and widest. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q49For Fraunhofer source rays, choose the correct diffraction statement.
Answer: A. parallel
Explanation: The correct result is parallel. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
Q50For Rayleigh criterion, choose the correct diffraction statement.
Answer: D. central maximum coincides with first minimum
Explanation: The correct result is central maximum coincides with first minimum. This is an original CBSE/NEET/JEE/IB/IGCSE/A-Level exam-style question.
16
40 Conceptual Questions
C1Why does diffraction increase when slit width decreases?
Answer: The angular spread scales approximately as λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C2Why does sound diffract more readily than visible light?
Answer: Sound wavelengths are comparable with everyday openings.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C3Why is light diffraction usually difficult to observe?
Answer: Visible wavelengths are extremely small compared with common apertures.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C4Why is the central maximum twice as wide?
Answer: Its bounding first minima lie at −λ/a and +λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C5Why is the central maximum brightest?
Answer: All aperture elements add in phase at θ = 0.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C6Why are secondary maxima weak?
Answer: Away from the center, most wavelet contributions cancel.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C7How does diffraction prove wave nature?
Answer: Bending and spreading require a wave description.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C8What is the role of Huygens wavelets?
Answer: Each aperture point emits a secondary wavelet.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C9Why are lenses used in Fraunhofer diffraction?
Answer: They create parallel incidence and focus equal-angle rays.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C10How does Fresnel diffraction differ from Fraunhofer diffraction?
Answer: Fresnel uses finite distances; Fraunhofer uses effectively infinite distances or lenses.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C11What fixes the first minimum?
Answer: The angular spread scales approximately as λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C12Why do two slit halves cancel at first minimum?
Answer: Sound wavelengths are comparable with everyday openings.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C13How are higher minima formed?
Answer: Visible wavelengths are extremely small compared with common apertures.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C14Why does λ increase central width?
Answer: Its bounding first minima lie at −λ/a and +λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C15Why does increasing D increase linear width?
Answer: All aperture elements add in phase at θ = 0.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C16Why does decreasing a broaden the pattern?
Answer: Away from the center, most wavelet contributions cancel.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C17Does frequency change during diffraction?
Answer: Bending and spreading require a wave description.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C18What is angular width of central maximum?
Answer: Each aperture point emits a secondary wavelet.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C19What is linear width of central maximum?
Answer: They create parallel incidence and focus equal-angle rays.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C20Where are minima located?
Answer: Fresnel uses finite distances; Fraunhofer uses effectively infinite distances or lenses.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C21What is the approximate secondary-maximum condition?
Answer: The angular spread scales approximately as λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C22Why is diffraction significant for a ≈ λ?
Answer: Sound wavelengths are comparable with everyday openings.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C23Can geometrical optics explain diffraction?
Answer: Visible wavelengths are extremely small compared with common apertures.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C24Why are edges not perfectly sharp?
Answer: Its bounding first minima lie at −λ/a and +λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C25What is a diffraction envelope?
Answer: All aperture elements add in phase at θ = 0.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C26Why is the central maximum symmetric?
Answer: Away from the center, most wavelet contributions cancel.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C27What does Rayleigh criterion mean?
Answer: Bending and spreading require a wave description.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C28What is limit of resolution?
Answer: Each aperture point emits a secondary wavelet.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C29How does aperture affect telescope resolution?
Answer: They create parallel incidence and focus equal-angle rays.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C30Why do large telescopes resolve better?
Answer: Fresnel uses finite distances; Fraunhofer uses effectively infinite distances or lenses.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C31How does microscope resolution depend on wavelength?
Answer: The angular spread scales approximately as λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C32Why is blue light better for resolution?
Answer: Sound wavelengths are comparable with everyday openings.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C33How does numerical aperture affect microscopy?
Answer: Visible wavelengths are extremely small compared with common apertures.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C34How do CDs produce colors?
Answer: Its bounding first minima lie at −λ/a and +λ/a.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C35How is X-ray diffraction useful?
Answer: All aperture elements add in phase at θ = 0.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C36Why are diffraction gratings used in spectroscopy?
Answer: Away from the center, most wavelet contributions cancel.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C37How does diffraction limit the human eye?
Answer: Bending and spreading require a wave description.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C38What happens when a ≫ λ?
Answer: Each aperture point emits a secondary wavelet.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C39What happens when a < λ?
Answer: They create parallel incidence and focus equal-angle rays.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
C40How is diffraction related to interference?
Answer: Fresnel uses finite distances; Fraunhofer uses effectively infinite distances or lenses.
Explanation: The result follows from superposition of Huygens wavelets across the aperture and the single-slit condition a sinθ = nλ.
17
Kumar Sir Exam Tips
Visibility
a ≈ λ → diffraction is prominent
Central Maximum
It is both widest and brightest.
Minima
a sinθ = nλ
Central Width
β₀ = 2Dλ/a
Wave Nature
Diffraction cannot be explained by ray optics alone.
Exam Focus
Minima and central-width numericals are frequent in NEET/JEE.