🛠️ JEE🧲 Physics

Wave Optics

2 solved JEE Physics previous year questions on Wave Optics, each with the correct answer and a full explanation.

Q1
In a Young's double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following? (i) Angular separation of the fringes. (ii) Fringe-width.
aBoth (i) and (ii) remain constant.
b(i) remains constant, but (ii) decreases.
c(i) remains constant, but (ii) increases.
dBoth (i) and (ii) increase.
✓ Correct answer: c) (i) remains constant, but (ii) increases.
ExplanationIn Young’s double-slit experiment (Y.D.S.E.), Fringe width (β): ​ where = wavelength of light, = distance between slits and screen, = distance between the two slits.Angular separation (θ): ​ Now: If the screen is moved away, increases.(i) Angular separation (θ) = ​ → independent of D, so remains constant.(ii) Fringe width (β) = → directly proportional to D, so increases. ✅ Correct Answer: ​
Q2
In a single-slit diffraction experiment, the width of the slit is halved. The width of the central maximum, in the diffraction pattern, will become :
ahalf
btwice
cfour times
done-fourth
✓ Correct answer: b) twice
ExplanationThe angular width ( ) of the central maximum in a single-slit diffraction pattern is given by the formula: If the original slit width a is halved, the new slit width a' is: The angular width doubles.Since the linear width of the central maximum on the screen is directly proportional to the angular width ( ), the linear width of the central maximum will also double.

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