Q.(a) A beam of light consisting of two wavelengths $400\ \text{nm}$ and $600\ \text{nm}$ is used to illuminate a single slit of width $1\ \text{mm}$. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed $1.5\ \text{m}$ from the slit.

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1. CBSE 2026Set 55/1/11 markMCQ
   Q.The shape of the interference fringes in Young's double-slit experiment, when the distance between the slit and the screen is very large as compared to the slit-separation, is nearly (A) straight (B) parabolic (C) circular (D) hyperbolic

   ›Reveal solutionSolution

   When the screen is very far from the slits compared to their separation, points of constant path difference lie on nearly straight lines parallel to the slits, making the fringes straight.

   The key to understanding fringe shape lies in recognizing what an interference fringe actually represents: it is the locus of all points on the screen where the path difference from the two slits is constant.

   In Young's double-slit experiment, we have two coherent sources $S_1$ and $S_2$ separated by distance $d$. A point $P$ on the screen at distance $D$ from the slits will show constructive or destructive interference depending on the path difference $\Delta = |S_2P - S_1P|$.

   For a bright fringe of order $n$, we need $\Delta = n\lambda$. The question is: what is the shape of the curve connecting all points $P$ that satisfy this condition?

   Geometry of path difference

   Consider a point $P$ on the screen at coordinates $(x, y)$ if we place the origin midway between the slits. The two slits are at positions roughly $(0, \pm d/2, 0)$ in 3D space, and the screen is at distance $D$ along the perpendicular.

   The exact path difference is:

   $\Delta = \sqrt{D^2 + (y - d/2)^2 + x^2} - \sqrt{D^2 + (y + d/2)^2 + x^2}$

   This is the general equation for a hyperbola in the $xy$-plane. So strictly speaking, fringes are hyperbolic curves.

   The far-field approximation

   Now comes the crucial condition: $D \gg d$ (screen distance much larger than slit separation).

   When $D$ is very large, we can use the binomial approximation. For the path from $S_2$ to $P$:

   $S_2P = D\sqrt{1 + \frac{(y-d/2)^2 + x^2}{D^2}} \approx D + \frac{(y-d/2)^2 + x^2}{2D}$

   Similarly for $S_1P$. The path difference becomes:

   $\Delta \approx \frac{(y-d/2)^2 - (y+d/2)^2}{2D} = \frac{-2yd}{2D} = -\frac{yd}{D}$

   (The $x^2$ terms cancel out.)

   For constant path difference $\Delta = n\lambda$, we get:

   $y = -\frac{n\lambda D}{d} = \text{constant}$ …

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2. CBSE 2026Set 55/2/11 markMCQ
   Q.Two statements are given – one labelled Assertion (A) and the other Reason (R). Select the correct answer from the codes below: (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false and (R) is also false. Assertion (A) : Light added to light can produce darkness. Reason (R) : When two coherent light waves interfere, there is darkness at the position of destructive interference.

   ›Reveal solutionSolution

   Interference of coherent light waves can indeed produce darkness at points of destructive interference; both statements are true and the reason correctly explains the assertion. The answer is (A).

   Why light can produce darkness

   The assertion sounds paradoxical at first—how can adding light to light create darkness? The resolution lies in understanding that light is a wave phenomenon, and waves don't simply add arithmetically in intensity. Instead, they superpose according to their phase relationship.

   When two coherent light waves (waves with a constant phase relationship) meet, their electric field amplitudes add vectorially. If they arrive in phase, the amplitudes reinforce (constructive interference, brighter light). If they arrive exactly out of phase—crest meeting trough—the amplitudes cancel (destructive interference), and the resultant intensity becomes zero or near-zero. This is darkness produced by adding light to light.

   The reason statement captures precisely this mechanism: destructive interference between coherent waves creates regions of darkness.

   Examining the statements

   1. Assertion (A): "Light added to light can produce darkness"

      This is experimentally verified in phenomena like Young's double-slit experiment, thin-film interference, and Newton's rings. At certain positions on the screen or observation plane, the intensity drops to zero despite light arriving from two sources. The statement is true.

   2. Reason (R): "When two coherent light waves interfere, there is darkness at the position of destructive interference"

      Destructive interference occurs when the path difference between two coherent waves is an odd multiple of half-wavelengths:

      $\Delta = \left(m + \frac{1}{2}\right)\lambda, \quad m = 0, 1, 2, \ldots$ …

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3. CBSE 2026Set V11 mark
   Q.The bending of light around the corners and entering into the geometric shadow region is called __________. Fill in the blank choosing the appropriate answer from the bracket: (photons, diffraction, polarity, monopoles, greater than unity, less than unity)

   ›Reveal solutionSolution

   diffraction …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Read the following passage carefully and answer the questions given below - Light entering in a dark room through a narrow gap under or around a closed door often appears to bend and spread. This is due to diffraction, the narrow gap acts like a slit, and light wave bends around the edges. This phenomenon is a small but clear demonstration of diffraction.

   (a) What would be the approximate size of sharp edge or opening compared to the wavelength of light for diffraction to be clearly observed?

   ›Reveal solutionSolution

   Diffraction is prominent only when the aperture/obstacle size is comparable to the wavelength of light.

   Diffraction effects become clearly noticeable only when the size of the slit/gap or obstacle is of the same order of magnitude as (comparable to) the wavelength of light used. If the opening is much larger than the wavelength, the bending is negligible and light appears to travel in straight lines (ray optics applies); as the opening size approaches the wavelength, diffraction (bending and spreading) becomes prominent — which is why the narrow …

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5. CBSE 2026Set ANNUAL1 mark
   Q.Read the following passage carefully and answer the questions given below - Light entering in a dark room through a narrow gap under or around a closed door often appears to bend and spread. This is due to diffraction, the narrow gap acts like a slit, and light wave bends around the edges. This phenomenon is a small but clear demonstration of diffraction.

   (b) Why can the diffraction not be explained by ray-optics?

   ›Reveal solutionSolution

   Ray optics assumes light always travels in straight lines and cannot describe the wave interference behind diffraction.

   Ray (geometrical) optics treats light purely as straight-line rays and predicts that an obstacle or slit will simply produce sharp-edged shadows/beams, with no bending at the edges. Diffraction, however, is fundamentally a wave phenomenon — it arises from the superposition (interference) of secondary wavelets originating from different points of the same wavefront as it passes an edge or narrow opening (Huygens–Fresnel principle). Since ray optics ignores the wave nature of …

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6. CBSE 2026Set A1 markMCQ
   Q.The phase difference φ is related to path difference λ by (A) (λ/π)φ (B) (π/λ)φ (C) (λ/2π)φ (D) (2π/λ)φ

   ›Reveal solutionSolution

   Phase difference = (2π/λ) × path difference.

   The fundamental relation between phase difference (Δϕ) and path difference (Δx) is

   $\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x.$

   …

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7. CBSE 2026Set A1 markMCQ
   Q.For destructive interference, the path difference should be equal to (A) nλ (B) (2n+1)λ/2 (C) zero (D) infinity

   ›Reveal solutionSolution

   Destructive interference occurs when the path difference is an odd multiple of λ/2.

   Two waves interfere destructively (cancel) when they arrive exactly out of phase, i.e. a phase difference of $\pi,3\pi,5\pi,\dots$. In terms of path difference this means an odd multiple of half a wavelength:

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.The displacement of water molecules at any instant on the surface of water at nodal lines is ______.

   ›Reveal solutionSolution

   Nodal lines are where two overlapping waves are always exactly out of phase, so their displacements cancel completely at every instant, leaving zero net displacement.

   When two coherent water-wave sources produce overlapping ripples, at points on a nodal line the crest of one wave always coincides with the trough of the other (path difference = odd multiple of half wavelength), so d …

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9. CBSE 2026Set ANNUAL1 mark
   Q.Define diffraction of light.

   ›Reveal solutionSolution

   Diffraction is the deviation of light from a straight-line path when it passes an obstacle or a narrow opening whose size is comparable to its wavelength.

   Diffraction of light is the phenomenon of bending of light waves around the corners of an obstacle or spreading of light after passing through a narrow slit/aperture, so that light appears in regions that would be a dark geometrical shadow according to simple ray optics. It becomes noticeable when the size of the obstacle/aperture is comparable to the wavelength of light, and is a direct consequence of the wave nature of light (a manifestatio …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The size of the obstacle for the diffraction of light should be

    (a) much larger than the wavelength of light

    (b) much smaller than the wavelength of light

    (c) of the order of wavelength of light

    (d) anything can happen

    ›Reveal solutionSolution

    Diffraction (bending of waves around obstacles) is significant only when the obstacle or slit size is comparable to the wavelength - much bigger or much smaller sizes don't show it clearly.

    Diffraction is the bending/spreading of waves as they pass an obstacle or through an aperture. This spreading is only prominent when the size of the obstacle/aperture (a) is of the SAME ORDER as the wavelength (lambda) of the wave, i.e. a ~ lambda. If the obstacle is much LARGER than the wavelength, the wave essentially travels in straight lines (geometrical shadow, negligible diffraction) - this is why we don't see visible light (wavelength ~500 nm) diffracting around everyday-s …

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11. CBSE 2026Set ANNUAL1 mark
    Q.What do you understand by the term 'diffraction of light'?

    ›Reveal solutionSolution

    Diffraction is the bending of light into the region that geometrical (ray) optics would call 'shadow' when it passes an obstacle or aperture.

    According to simple ray/geometrical optics, light travelling past an obstacle or through a slit should produce a sharp-edged shadow. In reality, being a wave, light bends slightly around the edges of the obstacle/aperture and spreads into what would otherwise be the shadow region, producing a pattern of bright and dark fringes near the edges. This bending and spreading of light waves around obstacles/apertures is called diffraction. It becomes prominent only when the size of the obstacle or aperture is comparable to the wavelength of light, …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.Select the correct option with respect to the figures given below:

    (a) Fig.

    (i) depicts diffraction pattern and Fig.

    (ii) depicts interference pattern

    (b) Both depict interference pattern

    (c) Both depict diffraction pattern

    (d) Fig.

    (i) depicts an interference pattern due to a double-slit and Fig.

    (ii) depicts a diffraction pattern due to a single-slit

    ›Reveal solutionSolution

    The key distinguishing feature between interference and diffraction intensity patterns is the relative heights of the fringes: interference from two narrow slits gives many fringes of roughly equal intensity, while diffraction from a single slit gives one strong central maximum with much weaker, rapidly falling secondary maxima.

    Distinguishing interference from diffraction patterns

    In Young's double-slit interference experiment, two coherent narrow sources produce a pattern of bright and dark fringes on the screen. Because the two slits are treated as (nearly) point/line sources of equal amplitude, the resulting bright fringes are all of comparable/roughly equal intensity across the region observed (there is a slowly-varying diffraction "envelope" from each slit's own finite width, but for the idealised interference pattern, the fringes near the centre appear as a series of similar-height peaks, evenly spaced by the fringe width $\beta=\lambda D/d$).

    In single-slit diffraction, light passing through one slit produces a pattern with:

    • one very bright central maximum (roughly twice as wide as the secondary maxima), and
    • a series of much fainter secondary maxima on either side, whose intensities fall off rapidly (roughly as $1/m^2$ or faster) with distance from the centre.

    Applying this to the two figures

    …

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