Q.A $5\ \text{cm}$ long pencil is placed along the principal axis of a concave mirror of focal length $20\ \text{cm}$ such that its nearest end is at a distance of $25\ \text{cm}$ from the mirror. Calculate the length of the image of the pencil.

Showing the 12 most recent of 40 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 2026Set ANNUAL1 markMCQ
   Q.A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. The location of the image is

   (a) formed at 6.67 cm behind the mirror.

   (b) formed at 67 cm behind the mirror.

   (c) formed at 70 cm same side of the mirror.

   (d) formed at 5.57 cm same side of the mirror.

   ›Reveal solutionSolution

   Using the mirror formula with the correct sign convention, the convex mirror forms a virtual image 6.67 cm behind the mirror.

   Given: Needle height $h=4.5$ cm, object distance $u = -12$ cm (object in front, so negative by convention), convex mirror so focal length $f = +15$ cm (behind the mirror, positive).

   Mirror formula:

   $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

   $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$

   Taking LCM (60): $\dfrac{1}{v} = \dfrac{4}{60} + \dfrac{5}{60} = \dfrac{9}{60} = \dfrac{3}{20}$

   $v = \frac{20}{3} = 6.67\ \text{cm}$ …

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7. CBSE 2025Set 55/4/11 markMCQ
   Q.The magnification produced by a spherical mirror is $-2.0$. The mirror used and the nature of the image formed will be: (A) Convex and virtual (B) Concave and real (C) Concave and virtual (D) Convex and real

   ›Reveal solutionSolution

   A magnification of $-2.0$ means the image is inverted (negative sign) and magnified (magnitude > 1). Only a concave mirror can produce an inverted, magnified image, and such an image is always real. So the mirror is concave and the image is real — option (B).

   Concept and Intuition

   The magnification $m$ of a spherical mirror tells you two things at once: the sign tells you orientation, and the magnitude tells you size.

   • If $m$ is positive, the image is virtual and erect (upright).
   • If $m$ is negative, the image is real and inverted (upside down).

   The magnitude $|m|$ tells you relative size:

   • $|m| > 1$ → image is magnified (larger than object)
   • $|m| < 1$ → image is diminished (smaller)
   • $|m| = 1$ → same size

   Here $m = -2.0$ means the image is inverted (negative) and twice as large as the object ($|m| = 2$).

   Now, which mirror can produce an inverted, magnified image? A convex mirror always gives a virtual, erect, and diminished image — so it can never produce a negative magnification. A concave mirror, however, can produce both real (inverted) and virtual (erect) images depending on where the object is placed. The real image from a concave mirror is always inverted, and when the object is between the centre of curvature and the focus, that real image is also magnified.

   So the only mirror that fits $m = -2.0$ is a concave mirror, and the image must be real.

   Step-by-step reasoning

   1. Interpret the sign of $m$

      $m = -2.0$ is negative. For spherical mirrors, a negative magnification always means the image is inverted relative to the object. An inverted image formed by a single mirror is always real (it can be projected on a screen). So the image is real.

   2. Interpret the magnitude of $m$

      $|m| = 2.0 > 1$, so the image is magnified — larger than the object.

   3. Eliminate convex mirror

      A convex mirror always produces a virtual, erect, and diminished image for any real object. That means $m$ is always positive and $|m| < 1$. Since our $m$ is negative and $|m| > 1$, a convex mirror is impossible. This eliminates options (A) and (D).

   4. Check concave mirror possibilities

      A concave mirror can produce:

      • A real, inverted, magnified image when the object is placed between $F$ and $C$ (focus and centre of curvature).
      • A virtual, erect, magnified image when the object is placed between $P$ and $F$ (pole and focus). In that case $m$ is positive.

      Since our $m$ is negative, the image cannot be virtual. So the only possibility is the real, inverted, magnified case — which is exactly what a concave mirror gives for an object between $F$ and $C$. …

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8. 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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9. 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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10. 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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11. CBSE 2025Set IMPROVEMENT1 markMCQ
    Q.Assertion (A): The radius of curvature of a concave mirror is 20 cm. If an object is placed in front of the mirror at a distance of 10 cm from its pole, its image is formed at infinity. Reason (R): The image of an object placed at the focus of a spherical mirror is formed at infinity. Select the correct option.

    (a) Both A and R are correct and R is the correct explanation of A.

    (b) Both A and R are correct but R is not the correct explanation of A.

    (c) A is correct but R is incorrect.

    (d) Both A and R are incorrect.

    ›Reveal solutionSolution

    Both statements are true, and the reason correctly explains why the assertion is true.

    For a concave mirror of radius of curvature $R = 20\,\text{cm}$, the focal length is $f = R/2 = 10\,\text{cm}$. In Assertion (A), the object is placed at a distance of 10 cm from the pole — exactly at the focus. Using the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$, when $u = f$, we get $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{f} = 0$, so $v \to \infty$ — the image is indeed formed at infinity. This is exactly the general principle stated in Reason (R): rays from an object placed at the f …

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12. 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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