Wave Optics JEE Main PYQs (2020–2026)

Solved Wave Optics PYQs

3 recent questions with answers. The full set of 131 is free to practise.

1. JEE Main 2026EasyInterference of Light Waves
   Q1

   In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $/3$ , where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda /6$ . The ratio of intensities at points $A$ and $B$ is $\_\_\_\_$ .

   • A3
   • B4
   • C1/3
   • D1/4

   Show answer & solution

   Correct answer: (C)

   The relation between phase difference $(\varphi )$ and path difference $(\Delta x)$ is:
   
   $$\varphi =\frac{2\pi }{\lambda }\Delta x$$
   
   For two waves of equal intensity $I_0$ , the resultant intensity I at a point is:
   
   $$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$$
   
   At point A the path difference is $\Delta x_A=\frac{\lambda }{3}$ .
   
   $${\varphi }_A=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$$
   
   So, the resultant intensity at A is,
   
   $$I_A=4I_0{\cos }^2\left(\frac{2\pi /3}{2}\right)=4I_0{\cos }^2\left(\frac{\pi }{3}\right)=4I_0{\left(\frac{1}{2}\right)}^2=I_0$$
   
   At point B the path difference is $\Delta x_B=\frac{\lambda }{6}$ .
   
   $${\varphi }_B=\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$$
   
   So, the resultant intensity at B is,
   
   $$I_B=4I_0{\cos }^2\left(\frac{\pi /3}{2}\right)=4I_0{\cos }^2\left(\frac{\pi }{6}\right)=4I_0{\left(\frac{\sqrt{3}}{2}\right)}^2=3I_0$$
   
   So, the ratio of intensities at A and B is;
   
   $$\frac{I_A}{I_B}=\frac{I_0}{3I_0}=\frac{1}{3}$$
   
   Thus, the correct option is (C).
2. JEE Main 2026EasyYoung's Double-Slit Experiment (YDSE)
   Q2

   In Young’s double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4\mu m$ . If the experiment is carried out in another medium having refractive index 1.2 , the fringe width will be $\_\_\_\_\mu m$ .

   • A1.2
   • B2
   • C2.4
   • D2.88

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   Correct answer: (B)

   In Young’s Double Slit Experiment (YDSE), the fringe width ( $\beta$ ) is given by :
   
   $$\beta =\frac{\lambda D}{d}$$
   
   Where, $\lambda$ is the wavelength of light used, D is the distance between the slits and the screen, and d is the distance between the two slits.
   
   When the experiment is moved from air/vacuum to a medium with a refractive index , the speed of light decreases, which in turn decreases the wavelength. The new wavelength ${\lambda }^{\prime }$ is:
   
   $${\lambda }^{\prime }=\frac{\lambda }{\mu }$$
   
   Since D and d are constant for the experimental setup, the new fringe width ${\beta }^{\prime }$ in the medium is :
   
   $${\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d}=\frac{\left(\frac{\lambda }{\mu }\right)D}{d}=\frac{1}{\mu }\left(\frac{\lambda D}{d}\right)$$
   
   Therefore ${\beta }^{\prime }=\frac{\beta }{\mu }$
   
   The original fringe width (when setup is in air) is $=2.4\mu m$ .
   
   The refractive index of the medium is $=1.2$ .
   
   Putting the values,
   
   $${\beta }^{\prime }=\frac{2.4}{1.2}=2.0\mu m$$
   
   Therefore, the correct option is (B).
3. JEE Main 2026MediumDiffraction of Light and Resolving Power
   Q3

   A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength 500 nm , at a resolution of $5\times 10^{-7}$ radian. The value of $R$ is
   
   $\_\_\_\_$ cm .

   • A61
   • B122
   • C244
   • D305

   Show answer & solution

   Correct answer: (B)

   The ability of an optical instrument to distinguish between two closely spaced objects is defined by its resolving power.
   
   According to the Rayleigh criterion, for a circular aperture of diameter R , the minimum angular separation ( $\Delta \theta$ ) that can be resolved is given by the formula :
   
   $$\Delta \theta =\frac{1.22\lambda }{R}$$
   
   Where $\Delta \theta$ is the angular resolution in radians, $\lambda$ is the wavelength of the incident light and $R$ is the diameter of the objective lens or mirror.
   
   We are given the wavelength of light as $\lambda =500nm=500\times 10^{-9}m=5\times 10^{-7}m$
   
   And angular resolution as $\Delta \theta =5\times 10^{-7}rad$
   
   $$R=\frac{1.22\lambda }{\Delta \theta }$$
   
   Substituting the values :
   
   $$R=\frac{1.22\times \left(5\times 10^{-7}\right)}{5\times 10^{-7}}$$
   
   $\Rightarrow R=1.22$ meters
   
   $$\Rightarrow R=1.22\times 100cm=122cm$$
   
   Therefore, the value of R is 122 cm .
   
   Thus, the correct option is (B).

Topic-wise question count

Where JEE Main has asked from Wave Optics, most frequent first.

• Young's Double-Slit Experiment (YDSE)50
• Interference of Light Waves31
• Polarization of Light21
• Diffraction of Light and Resolving Power18
• Huygens' Principle, Wavefronts and Doppler Effect11

Year-wise Wave Optics questions in JEE Main

Number of questions from this chapter in each year's paper(s).