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MedExams

Wave Optics

7 NEET-UG practice questions on Wave Optics, part of Physics. The first 7 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Easy

    Which phenomenon shows that light is a transverse wave?

    1. AInterference
    2. BPolarisation
    3. CDiffraction
    4. DRefraction
    Show answer

    Answer: B · Polarisation

    Only a transverse wave can have its oscillation confined to one plane, so polarisation shows that light is transverse. Interference and diffraction show only that it is a wave.

    • A: Longitudinal waves also interfere.
    • C: Longitudinal waves also diffract.
    • D: Refraction occurs for both kinds of wave.

    Remember this

    Sound in air, a longitudinal wave, interferes and diffracts but cannot be polarised.

  2. Question 2 · Medium

    Unpolarised light of intensity 64 W/m² passes through two ideal polaroids whose transmission axes are at 30° to each other. What intensity emerges from the second polaroid?

    1. A8 W/m²
    2. B24 W/m²
    3. C48 W/m²
    4. D27.7 W/m²
    Show answer

    Answer: B · 24 W/m²

    The first polaroid passes half the intensity, 32 W/m². Malus's law then gives I = 32 cos²30° = 32 × 3/4 = 24 W/m².

    • A: sin²30° was used in place of cos²30°.
    • C: The halving at the first polaroid was missed.
    • D: cos 30° was used without squaring.

    Remember this

    The first polaroid halves unpolarised light; each later polaroid multiplies by cos²θ of the angle between successive axes.

  3. Question 3 · Medium

    Parallel light of wavelength 500 nm falls normally on a single slit 0.1 mm wide. A screen is 1 m away. What is the width of the central maximum?

    1. A20 mm
    2. B1 mm
    3. C10 mm
    4. D5 mm
    Show answer

    Answer: C · 10 mm

    The first minima lie at y = ±λD/a = ±(5 × 10⁻⁷ × 1)/(10⁻⁴) = ±5 mm. The central maximum spans both, so its width is 2λD/a = 10 mm.

    • A: The first minimum was placed at 2λD/a and then doubled.
    • B: The slit width was read as 1 mm.
    • D: This is the distance from the centre to the first minimum only.

    Remember this

    Central maximum width is 2λD/a, twice every other fringe, and widens as the slit narrows.

  4. Question 4 · Medium

    A Young's double-slit set-up has slit separation 0.6 mm and screen distance 1.5 m, with light of wavelength 480 nm. How far apart are the third bright fringe on one side of the centre and the second dark fringe on the other side?

    1. A5.4 mm
    2. B3.6 mm
    3. C1.8 mm
    4. D6.0 mm
    Show answer

    Answer: A · 5.4 mm

    Fringe width β = λD/d = (480 × 10⁻⁹ × 1.5)/(0.6 × 10⁻³) = 1.2 mm. The third bright fringe is at 3β and the second dark fringe at 1.5β, on opposite sides. The separation is 4.5β = 5.4 mm.

    • B: This is the position of the third bright fringe alone.
    • C: This takes both fringes on the same side.
    • D: The second dark fringe was placed at 2β.

    Remember this

    The nth bright fringe lies at nβ and the nth dark fringe at (n - 1/2)β from the centre.

  5. Question 5 · Easy

    Plane polarised light of intensity I₀ falls on an analyser. θ is the angle between the plane of polarisation and the analyser's transmission axis. Which graph shows the transmitted intensity I against θ?

    Figure: MedExams
    Four graphs of transmitted intensity against angle from 0 to 180 degrees, labelled A to D, with 45 and 90 degrees marked.
    1. AGraph C
    2. BGraph B
    3. CGraph D
    4. DGraph A
    Show answer

    Answer: A · Graph C

    Malus's law gives I = I₀ cos² θ. So I = I₀ at 0° and 180°, zero at 90°, and I₀/2 at 45°. The guide lines at 45° pick out this curve.

    • B: Plots |cos θ|, which is the amplitude and not the intensity.
    • C: Fits unpolarised light, which gives I₀/2 at any angle.
    • D: Uses sin² θ, measuring θ from the wrong direction.

    Remember this

    Malus's law: I = I₀ cos² θ for already polarised light; unpolarised light passes one polaroid as I₀/2 at any angle.

  6. Question 6 · Hard

    The intensity pattern from two coherent slits of unequal widths is plotted here. Find the ratio of the intensities from the two slits, and the intensity at a point where the path difference is λ/4.

    Figure: MedExams
    Intensity against position for a two-slit pattern, with the maximum and minimum intensities marked in terms of I0.
    1. A4 : 1 and 4.5I₀
    2. B9 : 1 and 10I₀
    3. C4 : 1 and 5I₀
    4. D2 : 1 and 5I₀

    Answer and explanation in LearnFree account

  7. Question 7 · Medium

    The fringe pattern on the screen of a Young's double-slit set-up is shown, with a measured length marked between the centres of two bright fringes. What is the wavelength of the light?

    Figure: MedExams
    A strip of bright and dark fringes with a 9.0 mm length marked between two bright fringes, and slit values written above.
    1. A500 nm
    2. B600 nm
    3. C750 nm
    4. D300 nm

    Answer and explanation in LearnFree account

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