Wave Optics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Wave Optics, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 15 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   In a single slit diffraction experiment, a slit of width $(0.016 \pm 0.002)$ mm is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and first minimum is measured to be $(2^\circ \pm 40')$. The value of the fractional error in the measurement of wavelength is: [Given: $\sin(2^\circ) = 0.035$]
   Show answer & solution

   Answer: 0.46

   For a single slit diffraction, the condition for the first minimum is given by: $a \sin \theta = \lambda$ Taking the natural logarithm on both sides, we get: $\ln \lambda = \ln a + \ln(\sin \theta)$ Differentiating to find the maximum fractional error: $\dfrac{\Delta \lambda}{\lambda} = \dfrac{\Delta a}{a} + \cot \theta \Delta \theta$ Since $\theta = 2^\circ$ is very small, $\cos \theta \approx 1$, which gives $\cot \theta \approx \dfrac{1}{\sin \theta}$. Given $\sin(2^\circ) = 0.035$, we can approximate the angle in radians as $\theta \approx \sin \theta = 0.035$ rad. The error in the angle is $\Delta \theta = 40' = \left(\dfrac{40}{60}\right)^\circ = \left(\dfrac{2}{3}\right)^\circ$. Converting $\Delta \theta$ into radians: $\Delta \theta = \dfrac{2}{3} \times \left(\dfrac{0.035}{2}\right) = \dfrac{0.035}{3}$ rad The fractional error in the angular term is: $\cot \theta \Delta \theta \approx \dfrac{\Delta \theta}{\sin \theta} = \dfrac{\dfrac{0.035}{3}}{0.035} = \dfrac{1}{3}$ The fractional error in the slit width is: $\dfrac{\Delta a}{a} = \dfrac{0.002}{0.016} = \dfrac{1}{8}$ Substituting these values into the error equation: $\dfrac{\Delta \lambda}{\lambda} = \dfrac{1}{8} + \dfrac{1}{3} = \dfrac{3 + 8}{24} = \dfrac{11}{24}$ Answer: $0.46$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   In the List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the correct match between the entries in List-I to those in List-II. List-IList-II(P) Colorful sky in north polar region (Aurora Borealis)(1) Dispersion and reflection(Q) Partially polarized sun light(2) Total internal reflection(R) Rainbow(3) Diffraction(S) Dark and bright fringes(4) Scattering of light by molecules in the atmosphere(5) Emission of radiation from oxygen and nitrogen atoms excited by charged particles
   1. A.P$\rightarrow$5, Q$\rightarrow$4, R$\rightarrow$1, S$\rightarrow$3
   2. B.P$\rightarrow$4, Q$\rightarrow$2, R$\rightarrow$1, S$\rightarrow$3
   3. C.P$\rightarrow$4, Q$\rightarrow$1, R$\rightarrow$2, S$\rightarrow$3
   4. D.P$\rightarrow$5, Q$\rightarrow$4, R$\rightarrow$1, S$\rightarrow$2
   Show answer & solution

   Answer: (A)

   The colorful sky in the north polar region, known as Aurora Borealis, is caused by the emission of radiation from oxygen and nitrogen atoms in the upper atmosphere when they are excited by charged particles from the solar wind. Thus, P $\rightarrow$ 5. Sunlight scattered by molecules in the atmosphere becomes partially polarized in a direction perpendicular to the incident light. Thus, Q $\rightarrow$ 4. A rainbow is formed due to the dispersion of sunlight and its internal reflection by water droplets in the atmosphere. Thus, R $\rightarrow$ 1. Dark and bright fringes are the characteristic patterns formed due to the superposition of light waves, which occurs in phenomena like interference and diffraction. Thus, S $\rightarrow$ 3. The correct matching is P $\rightarrow$ 5, Q $\rightarrow$ 4, R $\rightarrow$ 1, S $\rightarrow$ 3. Answer: P$\rightarrow$5, Q$\rightarrow$4, R$\rightarrow$1, S$\rightarrow$3
3. Q3JEE Advanced Adv 2025 (Paper 1)
   A single slit diffraction experiment is performed to determine the slit width using the equation, $\frac{b d}{D}=m \lambda$, where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m^{\text {th }}$ diffraction maximum and the central maximum, and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of 1 cm and 1 mm, respectively. The values of $\lambda$ and $m$ are known precisely to be 600 nm and 3, respectively. The maximum absolute error (in $\mu \mathrm{m}$) in the value of $b$ estimated using the diffraction maximum that occurs for $m=3$ with $d=5 \mathrm{~mm}$ and $D=1 \mathrm{~m}$ is ______.
   Show answer & solution

   Answer: 94.5

   Original question asked about absolute error which can be both minimum or maximum, we added word maximum to make it clear for students $\begin{aligned} & \mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=360 \mu \mathrm{~m} \\ & \mathrm{~b}_{\max }=\frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \mu \mathrm{~m}=454.5 \mu \mathrm{~m} \\ & \mathrm{~b}_{\min }=\frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m}\end{aligned}$ Maximum value of b gives error, $\Delta \mathrm{b}_1=94.5 \mu \mathrm{~m}$ Minimum value of $b$ gives error, $\Delta b_2=63 \mu \mathrm{~m}$ $\therefore$ We always report the largest error, hence correct answer should be $94.5 \mu \mathrm{~m}$
4. Q4JEE Advanced Adv 2024 (Paper 1)
   A point source $\mathrm{S}$ emits unpolarized light uniformly in all directions. At two points $\mathrm{A}$ and $\mathrm{B}$, the ratio $r=I_A / I_B$ of the intensities of light is 2. If a set of two polaroids having $45^{\circ}$ angle between their pass-axes is placed just before point $\mathrm{B}$, then the new value of $r$ will be ______
   Show answer & solution

   Answer: 8

   $I \propto \frac{1}{l^2}$ Where I: distance from point source. $\begin{aligned} & \Rightarrow \frac{I_A}{I_B}=\frac{I_B^2}{I_A^2}=2 \\ & \Rightarrow I_B=\sqrt{2} I_A . \end{aligned}$ Also, due to polaroids: $\begin{aligned} & I_B^{\prime}=\frac{I_B}{2} \cos ^2 45^{\circ}=\frac{I_B}{4} \\ & \Rightarrow I_B^{\prime}=\frac{I_B}{4} \ldots \text { (2) } \\ & \Rightarrow \text { Ratio becomes } 4 \text { times. } \\ & \Rightarrow r_{\text {new }}=8 \end{aligned}$
5. Q5JEE Advanced Adv 2014 (Paper 1)
   A light source, which emits two wavelengths ${\lambda }_{1}=400nm$ and ${\lambda }_{2}=600nm$ , is used in a Young's double slit experiment. If recorded fringe widths for ${\lambda }_{1}$ and ${\lambda }_{2}$ are ${\beta }_{1}$ and ${\beta }_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are ${m}_{1}$ and ${m}_{2}$ , respectively, then
   1. A.${\beta }_{2}\gt {\beta }_{1}$
   2. B.${m}_{1}\gt {m}_{2}$
   3. C.From the central maximum, ${3}^{rd}$ maximum of ${\lambda }_{2}$ overlaps with ${5}^{th}$ minimum of ${\lambda }_{1}$
   4. D.The angular separation of fringes for ${\lambda }_{1}$ is greater than ${\lambda }_{2}$
   Show answer & solution

   Answer: A,B,C

   $\beta =\dfrac{D\lambda }{d}$ $∵{\lambda }_{2}\gt {\lambda }_{1}\Rightarrow {\beta }_{2}\gt {\beta }_{1}$ Also ${m}_{1}{\beta }_{1}={m}_{2}{\beta }_{2}\Rightarrow {m}_{1}\gt {m}_{2}$ Also $3\left(\dfrac{D}{d}\right)\left(600nm\right)=\left(2\times 5-1\right)\left(\dfrac{D}{2d}\right)400nm$ Angular width $\theta =\dfrac{\lambda }{d}$