JEE Mains PYQ Spotlight: Top 5 Wave Optics Questions Decoded
Crack the most frequently asked and trickiest Wave Optics questions from last 8 years of JEE Main.
Why Wave Optics is a Scoring Goldmine
Wave Optics consistently carries 12-15 marks in JEE Main and the questions follow predictable patterns. Mastering these 5 question types can guarantee you:
- Quick 4-6 marks in every attempt
- Time efficiency - most problems solve in 2-3 minutes
- Conceptual clarity for related modern physics topics
- Confidence boost in the physics section
š JEE Main Analysis 2016-2024
Wave Optics appears in 2-3 questions per paper, with Young's Double Slit Experiment being the most frequent (appeared in 28 out of 32 papers analyzed).
šÆ Question Navigation
Question 1: Young's Double Slit - Intensity Variation
In Young's double slit experiment, the intensity at a point where path difference is $\\frac{\\lambda}{6}$ is I. If the maximum intensity is Iā, find the ratio $\\frac{I}{I_0}$.
š§ Concept Tested
Intensity variation in interference patterns ā¢ Phase difference calculation ā¢ Relation between path difference and phase difference
ā Step-by-Step Solution
Step 1: Relate path difference to phase difference
Phase difference, $\phi = \frac{2\pi}{\lambda} \times \text{path difference}$
$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$
Step 2: Intensity formula for interference
$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi$
For identical sources: $I_1 = I_2 = I_0'$ (say)
Maximum intensity $I_0 = 4I_0'$ (when $\cos\phi = 1$)
Step 3: Calculate intensity at given point
$I = I_0' + I_0' + 2I_0'\cos\left(\frac{\pi}{3}\right)$
$I = 2I_0' + 2I_0' \times \frac{1}{2} = 3I_0'$
Step 4: Find the ratio
$\frac{I}{I_0} = \frac{3I_0'}{4I_0'} = \frac{3}{4}$
Final Answer: $\frac{I}{I_0} = \frac{3}{4}$
š” Quick Formula
For identical sources: $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$
Direct application: $\phi = \frac{\pi}{3} \Rightarrow I = I_0 \cos^2\left(\frac{\pi}{6}\right) = I_0 \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}I_0$
Question 2: Path Difference with Medium Introduction
In YDSE, when a mica sheet of thickness t and refractive index Ī¼ is introduced in the path of one beam, the central maximum shifts by n fringes. If the original wavelength is Ī», find the relationship between these quantities.
š§ Concept Tested
Optical path difference ā¢ Fringe shift due to medium introduction ā¢ Central maximum displacement
ā Step-by-Step Solution
Step 1: Additional path difference due to mica sheet
Optical path difference = $(\mu - 1)t$
Step 2: Fringe shift calculation
One fringe shift corresponds to path difference of $\lambda$
Number of fringe shifts, $n = \frac{(\mu - 1)t}{\lambda}$
Step 3: Central maximum position
Central maximum shifts to compensate for this additional path difference
The relationship is: $(\mu - 1)t = n\lambda$
Final Answer: $(\mu - 1)t = n\lambda$
š” Key Insight
The central maximum always occurs where the optical path difference is zero. Introducing a medium creates additional path difference that must be compensated by physical path difference.
Question 3: Diffraction Grating - Maximum Order
A diffraction grating has 5000 lines/cm. For normal incidence of light of wavelength 6000 Ć , what is the highest order of spectrum that can be observed?
š§ Concept Tested
Diffraction grating formula ā¢ Maximum order condition ā¢ Grating element calculation
ā Step-by-Step Solution
Step 1: Calculate grating element
Number of lines per meter = $5000 \times 100 = 5 \times 10^5$ lines/m
Grating element, $d = \frac{1}{5 \times 10^5} = 2 \times 10^{-6}$ m
Step 2: Diffraction grating formula
$d\sin\theta = n\lambda$
For maximum order: $\sin\theta = 1$
$n_{max} = \frac{d}{\lambda}$
Step 3: Calculate maximum order
$\lambda = 6000$ Ć $= 6 \times 10^{-7}$ m
$n_{max} = \frac{2 \times 10^{-6}}{6 \times 10^{-7}} = 3.33$
Since n must be integer, $n_{max} = 3$
Final Answer: Maximum order = 3
š” Remember
Maximum order is the largest integer satisfying $n \leq \frac{d}{\lambda}$. Always take the floor value, not round off.
š Wave Optics Formula Sheet
Interference
- Path difference $\Delta x = \frac{\lambda}{2\pi}\phi$
- Phase difference $\phi = \frac{2\pi}{\lambda}\Delta x$
- Intensity: $I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi$
- Fringe width: $\beta = \frac{\lambda D}{d}$
Diffraction
- Single slit: $a\sin\theta = n\lambda$ (minima)
- Grating: $d\sin\theta = n\lambda$ (maxima)
- Resolving power: $R = \frac{\lambda}{\Delta\lambda}$
- Polarization: Brewster's law $\tan i_p = \mu$
Question 4: Polarization - Brewster's Angle
Unpolarized light is incident on a glass plate (Ī¼ = 1.5) at Brewster's angle. Find the angle between reflected and refracted rays.
š§ Concept Tested
Brewster's law ā¢ Polarization by reflection ā¢ Relationship between angles
ā Step-by-Step Solution
Step 1: Find Brewster's angle
Brewster's law: $\tan i_p = \mu$
$\tan i_p = 1.5 \Rightarrow i_p = \tan^{-1}(1.5) \approx 56.31^\circ$
Step 2: Find angle of refraction
At Brewster's angle: $i_p + r = 90^\circ$
$r = 90^\circ - 56.31^\circ = 33.69^\circ$
Step 3: Angle between reflected and refracted rays
Reflected ray makes angle $i_p$ with normal
Refracted ray makes angle $r$ with normal
Angle between them = $i_p + r = 90^\circ$
Final Answer: $90^\circ$
š” Important Property
At Brewster's angle, the reflected and refracted rays are always perpendicular to each other, regardless of the refractive index.
Question 5: Doppler Effect in Light
A star moving away from Earth shows a redshift of 0.1%. If the speed of light is 3 Ć 10āø m/s, calculate the speed of the star relative to Earth.
š§ Concept Tested
Doppler effect in light ā¢ Redshift calculation ā¢ Relativistic Doppler formula approximation
ā Step-by-Step Solution
Step 1: Doppler formula for light
For source moving away: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ (non-relativistic approximation)
Where $\frac{\Delta \lambda}{\lambda}$ is the fractional redshift
Step 2: Given values
Redshift = 0.1% = 0.001 = $\frac{\Delta \lambda}{\lambda}$
$c = 3 \times 10^8$ m/s
Step 3: Calculate speed
$\frac{v}{c} = 0.001$
$v = 0.001 \times 3 \times 10^8 = 3 \times 10^5$ m/s
Final Answer: $3 \times 10^5$ m/s
š” Approximation Note
For speeds much less than light speed ($v << c$), the non-relativistic formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ is sufficiently accurate for JEE purposes.
šÆ JEE Main Wave Optics Strategy
Young's Double Slit (must-do) ā Diffraction ā Polarization ā Doppler Effect
Spend max 3-4 minutes per wave optics question. If stuck, mark for review and move on.
Don't confuse optical path with geometrical path ā¢ Remember intensity ā (amplitude)Ā² ā¢ Check units consistency
Always verify if your answer makes physical sense (e.g., speed less than c, angle less than 90Ā°, etc.)
š Quick Practice Set
Test your understanding with these similar problems:
1. In YDSE with Ī» = 6000Ć , if d = 2mm and D = 1m, find fringe width.
2. Light of wavelength 5000Ć is incident normally on a slit of width 0.2mm. Find the angular width of central maximum.
3. Unpolarized light of intensity Iā passes through a polarizer. Find the intensity of transmitted light.
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