Physics · chapter PYQs

Wave Optics, on the exam screen.

3 questions from NEET 2026, NEET 2020, 3 minutes, marked +4 / −1. See them with solutions.

Wave Optics · PYQs

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Question 1 · NEET 2026 · Q17 · +4 / −1
In Young's double slit experiment, using monochromatic light of wavelength $\lambda$ , the intensity of light at a point on the screen where the path difference is $\lambda$ , is $K$ units. The intensity of light at a point where the path difference is $\dfrac{\lambda}{3}$ will be
Answer and explanation
Answer: (1)
The interference intensity follows $I = I_0\cos^2\left(\dfrac{\Delta\phi}{2}\right)$ . A path difference of $\lambda$ gives $\Delta\phi = 2\pi$ , so $I = I_0 = K$ , i.e. $I_0 = K$ . For a path difference $\lambda/3$ , $\Delta\phi = \dfrac{2\pi}{3}$ , giving $I = I_0\cos^2\left(\dfrac{\pi}{3}\right) = \dfrac{I_0}{4} = \dfrac{K}{4}$ (option (1)). The other choices ignore the $\cos^2$ dependence on phase.
Question 2 · NEET 2026 · Q20 · +4 / −1
In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. Evaluate the following statements and choose the correct answer:

Statement I: As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
Statement II: Diffraction and interference are characteristics exhibited only by light waves.
Answer and explanation
Answer: (3)
Interference and diffraction only redistribute light energy, so darkening of one region is compensated by brightening of another, keeping total energy unchanged. This makes statement I consistent with conservation of energy, i.e. Statement I is true. However, interference and diffraction are general wave phenomena shown by sound, water and other waves, not only light, so statement II is false. Hence 'I true, II false' (option (3)).
Question 3 · NEET 2020 · Q172 · +4 / −1
In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :
Answer and explanation
Answer: (2)
The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$ , where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the separation between the slits. If $d$ is halved and $D$ is doubled, $\beta$ becomes $\beta' = \frac{\lambda (2D)}{\frac{d}{2}} = 4 \beta$ . Therefore, the fringe width becomes four times. Chapter: Wave Optics.