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Additional Exercises · 10.17

Q.Answer the following questions:

(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m7\ \text{m} partition wall in a room 10 m10\ \text{m} high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Telangana TsbieTextbookSubjective· 5mImportance★★★★★est
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  1. Doubling the slit width halves the central band's angular width and quadruples its intensity.
  2. The double-slit interference pattern is modulated (enveloped) by the single-slit diffraction pattern of each individual slit.
  3. The bright central spot behind a circular obstacle arises because diffracted waves from every point on the obstacle's rim travel equal distances to the shadow's centre and arrive in phase.
  4. Sound's much longer wavelength diffracts around the partition easily, while light's tiny wavelength does not. (e) Ray optics remains valid because ordinary apertures are far larger than the wavelength of light, so diffraction spreading is negligible over the relevant distances.
  1. Doubling the slit width In single-slit diffraction, the angular half-width of the central maximum is θ≈λ/a\theta \approx \lambda/a. Doubling the slit width a→2aa \to 2a halves this angular width, so the central diffraction band becomes narrower (half its original angular size). At the same time, since twice as much light passes through the wider slit, the amplitude at the centre (which adds up constructively from every point across the slit) doubles, and since intensity ∝(amplitude)2\propto (\text{amplitude})^2, the intensity at the centre becomes four times as large.
  2. Diffraction and the double-slit interference pattern The observed double-slit pattern is not pure interference fringes of uniform brightness — it is the fine, closely-spaced interference fringes (set by the slit separation dd) modulated by the broader single-slit diffraction envelope (set by each slit's own width aa). The diffraction pattern of a single slit acts as an envelope that limits how bright each interference fringe can be — fringes near a diffraction minimum are strongly suppressed, and no interference fringes are visible outside the central diffraction envelope.
  3. The bright spot at the centre of a circular obstacle's shadow Every point on the rim (edge) of a small circular obstacle is, by symmetry, exactly the same distance from the centre point of the shadow directly behind it. Light diffracted from all points around the rim therefore travels equal path lengths to reach that centre point, arriving there exactly in phase, and interferes constructively — producing a bright spot at the very centre of the geometrical shadow. (This is the famous Poisson's spot, first predicted as a seemingly absurd consequence of Fresnel's wave theory and then experimentally confirmed by Arago — becoming one of the strongest pieces of evidence for the wave nature of light.)
  4. Why the partition blocks light but not sound Appreciable diffraction (bending around an obstacle) occurs only when the size of the obstacle is comparable to the wavelength of the wave. Audible sound has wavelengths of the order of centimetres to metres — comparable to the 7 m7\ \text{m} partition wall — so sound diffracts around it easily and reaches the other side. Visible light has a wavelength of only about 5×10−7 m5\times10^{-7}\ \text{m}, many orders of magnitude smaller than the 7 m7\ \text{m} wall, so light diffracts negligibly around it and travels essentially in straight lines — the students can hear each other but cannot see each other. …

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