Q.(a)

Showing the 12 most recent of 32 on this concept.

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 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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4. 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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5. 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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6. CBSE 2025Set 55/5/11 markMCQ
   Q.A compound microscope has an objective and an eyepiece of focal lengths $f_{o}$ and $f_{e}$ respectively. To obtain a large magnification of a small object, the microscope should have: (A) $f_{o}$ and $f_{e}$ small, and $f_{e}>f_{o}$ (B) $f_{o}$ and $f_{e}$ small, and $f_{o}>f_{e}$ (C) $f_{o}$ and $f_{e}$ large, and $f_{e}>f_{o}$ (D) $f_{o}$ and $f_{e}$ large, and $f_{o}>f_{e}$

   ›Reveal solutionSolution

   A compound microscope achieves high magnification when both lenses have short focal lengths, with the objective's focal length shorter than the eyepiece's: $f_o$ and $f_e$ small, and $f_e > f_o$. The correct option is (A).

   Why focal lengths matter for magnification

   A compound microscope uses two converging lenses in tandem. The objective forms a real, magnified image of the tiny object, and the eyepiece acts as a magnifying glass to view that intermediate image. The total magnification is the product of the magnifications produced by each lens, so to maximize it we need to understand how each focal length enters the formula.

   The objective's magnification depends on how far the real image forms relative to the object distance. For a small object placed just beyond the focal point of the objective, the image distance $v_o$ is much larger than the object distance $u_o$, giving magnification $m_o = \frac{v_o}{u_o}$. In the standard microscope setup, the object is very close to the focus, so $u_o \approx f_o$, and the image forms near the other end of the tube at distance $L$ (the tube length). This gives:

   $m_o \approx \frac{L}{f_o}$

   The eyepiece magnifies the intermediate image like a simple magnifying glass. When the final image is at the near point $D$ (typically 25 cm), the angular magnification is:

   $m_e = 1 + \frac{D}{f_e}$

   For large $D$ compared to $f_e$, this simplifies to $m_e \approx \frac{D}{f_e}$.

   $M = m_o \times m_e \approx \frac{L}{f_o} \times \frac{D}{f_e} = \frac{LD}{f_o f_e}$

   Step-by-step analysis

   1. Total magnification is inversely proportional to both focal lengths. From the formula above, $M \propto \frac{1}{f_o f_e}$. To maximize $M$, we need both $f_o$ and $f_e$ to be as small as possible. This immediately rules out options (C) and (D), which suggest large focal lengths.

   2. Compare the two focal lengths. Now we must decide between options (A) and (B): should $f_e > f_o$, or $f_o > f_e$?

   3. Physical constraints of microscope design. In a practical compound microscope, the objective must form a real image at a reasonable distance (the tube length $L$, typically 15–20 cm). If $f_o$ were too large, the object would need to be placed far from the lens, defeating the purpose of examining a small specimen up close. The objective typically has $f_o$ in the range of a few millimetres to about 2 cm. …

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7. CBSE 2025Set 55/5/11 markMCQ
   Q.Two coherent light waves, each having amplitude $a$, superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between: (A) 0 and $2a^{2}$ (B) 0 and $4a^{2}$ (C) $a^{2}$ and $2a^{2}$ (D) $2a^{2}$ and $4a^{2}$

   ›Reveal solutionSolution

   When two coherent waves of equal amplitude $a$ interfere, the resultant amplitude varies from $0$ (destructive) to $2a$ (constructive); since intensity is proportional to the square of amplitude, the intensity range is 0 to $4a^2$.

   The heart of this problem lies in understanding how wave superposition affects intensity. When two coherent waves meet, they don't simply add their intensities — instead, their amplitudes add vectorially, and the resulting intensity depends on the square of this net amplitude.

   For light waves, intensity $I$ is proportional to the square of the amplitude: $I \propto A^2$. If we set the proportionality constant to unity for simplicity (which is standard when comparing relative intensities), then $I = A^2$.

   Now let's trace what happens when two coherent waves, each with amplitude $a$, interfere.

   The amplitude addition principle

   At any point on the screen, the two waves arrive with some phase difference $\delta$ (which depends on the path difference). The resultant amplitude is found by vector addition:

   $A_{\text{net}} = a_1 + a_2$

   where $a_1$ and $a_2$ are the individual wave amplitudes treated as phasors. For two waves of equal amplitude $a$ with phase difference $\delta$:

   $A_{\text{net}} = \sqrt{a^2 + a^2 + 2a \cdot a \cos\delta} = a\sqrt{2(1 + \cos\delta)}$

   Using the identity $1 + \cos\delta = 2\cos^2(\delta/2)$:

   $A_{\text{net}} = 2a\left|\cos\frac{\delta}{2}\right|$

   Finding the intensity extremes

   1. Maximum amplitude (constructive interference): When $\delta = 0, 2\pi, 4\pi, \ldots$ (waves in phase), we have $\cos(\delta/2) = 1$, so:

   $A_{\max} = 2a$

   The maximum intensity is:

   $I_{\max} = A_{\max}^2 = (2a)^2 = 4a^2$ …

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8. CBSE 2025Set 55/6/11 markMCQ
   Q.Two coherent waves, each of intensity $I_0$, produce interference pattern on a screen. The average intensity of light on the screen is: (A) zero (B) $I_0$ (C) $2I_0$ (D) $4I_0$

   ›Reveal solutionSolution

   Interference redistributes light energy across the screen but cannot create or destroy it, so the average intensity equals the sum of the two individual intensities: $I_0 + I_0 = 2I_0$. The correct option is (C).

   When two coherent waves meet, they produce bright and dark fringes. At some points they add constructively (bright), at others destructively (dark). The question asks for the average intensity over the whole screen — not the maximum or minimum at any particular point.

   The key insight is energy conservation. The two sources together deliver a fixed amount of energy to the screen. Interference only redistributes this energy — concentrating it in bright fringes and depleting it in dark ones — it cannot create or destroy energy. So the average over the pattern must equal what the two waves would deliver independently.

   Let's confirm this with the intensity formula.

   1. Write the resultant intensity at a point. For two coherent waves of intensity $I_0$ each, meeting with phase difference $\delta$:

   $I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\,\cos\delta = 2I_0(1 + \cos\delta)$

   This varies from $I_{\text{max}} = 4I_0$ (at $\delta = 0, 2\pi, \ldots$) down to $I_{\text{min}} = 0$ (at $\delta = \pi, 3\pi, \ldots$).

   2. Average over the pattern. As you move across the screen, the phase difference $\delta$ sweeps uniformly through all values from $0$ to $2\pi$:

   $\langle I \rangle = 2I_0\left(1 + \langle\cos\delta\rangle\right)$

   3. Evaluate the average of cosine.

      Over a complete cycle, $\langle\cos\delta\rangle = 0$.

   4. Conclude.

      $\langle I \rangle = 2I_0(1 + 0) = 2I_0$ …

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9. CBSE 2025Set X11 markMCQ
   Q.Which one of the following statements is WRONG about interference of light?

   (a) Light waves of same wavelength coming from two independent sources can be coherent and can produce interference

   (b) When the path difference between two interfering waves in $n\lambda$, bright fringe is produced (Here $n = 0, 1, 2, \ldots$ and $\lambda$ is the wavelength of light)

   (c) When the phase difference between two interfering waves is $(2n+1)\pi$, dark fringe is produced (Here $n = 0, 1, 2, \ldots$)

   (d) In Young's double slit experiment, dark and bright fringes are equally spaced

   ›Reveal solutionSolution

   (a) — this is the WRONG statement. Two independent sources cannot maintain a constant phase relationship, so they are not coherent and cannot produce a stable interference pattern; coherent sour …

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10. CBSE 2025Set D1 markMCQ
    Q.Two light waves of equal amplitude and equal wavelengths are superimposed. The amplitude of the resultant wave will be maximum when phase difference between them is (A) zero (B) π/4 (C) π/2 (D) π

    ›Reveal solutionSolution

    Two equal-amplitude waves add to a maximum when they are in phase, i.e. phase difference = 0.

    For two waves of amplitude a with phase difference φ, the resultant amplitude is

    A = √(a² + a² + 2a·a·cosφ) = 2a·cos(φ/2)

    …

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11. CBSE 2025Set D1 markMCQ
    Q.The bubble of soap appears coloured due to (A) diffraction (B) polarization (C) interference (D) reflection

    ›Reveal solutionSolution

    Thin-film interference between the two reflected rays gives colour, since which wavelength interferes constructively depends on the film thickness.

    A soap film is a very thin layer of liquid. Light reflected from its front surface interferes with light reflected from its back surface (thin-film interference).

    …

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12. CBSE 2025Set A1 mark
    Q.Fill in the blank with appropriate word: ______ of light shows redistribution of energy.

    ›Reveal solutionSolution

    Interference of light merely redistributes energy between bright and dark regions; it does not create or destroy energy.

    When two coherent light waves superpose, they produce a pattern of alternating bright and dark fringes. At points of constructive interference, the resultant intensity is greater than the sum of individual intensities (extra brightness), while at points of destructive interference the resultant intensity is less (often close to zero). Averaged over the whole pattern, however, the total energy is exactly conserved — …

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