A passenger in an aeroplane shall (a) never see a rainbow. (b) may see a primary and a secondary rainbow as concentric circles. (c) may see a primary and a secondary rainbow as concentric arcs. (d) shall never see a secondary rainbow.

A passenger in an aeroplane shall (a) never see a rainbow. (b) may see a primary and a secondary rainbow as co

NCERT Exemplar · Q4

Q. A passenger in an aeroplane shall

(a) never see a rainbow.

(b) may see a primary and a secondary rainbow as concentric circles.

(d) shall never see a secondary rainbow.

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Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

Imagine you're standing in front of a concave mirror — the kind that makes your face look bigger when you're close, but flips everything upside down when you step far back. That change isn't magic; it's geometry. The spherical mirror equation is the single relationship that predicts exactly where an image will form, and whether it's real or virtual, for any spherical mirror.

The Core Idea

Every point on an object sends out light rays in all directions. A mirror redirects those rays. The mirror equation tells you: given the mirror's curvature and the object's distance, where will those rays meet again (or appear to meet)?

There are only three quantities you need:

$u$ — object distance (from the mirror's pole)
$v$ — image distance (from the mirror's pole)
$f$ — focal length (a property of the mirror's curvature)

The equation is:

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

The power is in the sign convention, because every distance can point in one of two directions.

The Sign Convention (New Cartesian Sign Convention)

This is where most students slip. The equation works for all spherical mirrors — concave and convex — only if you follow the convention used throughout NCERT and CBSE:

All distances are measured from the mirror's pole.
The incident light is taken to travel left to right, so distances measured in that same direction (to the right) are positive, and distances measured against it (to the left) are negative.
Heights above the principal axis are positive; heights below are negative.

Because a real object is always placed in front of the mirror (to the left, where the incident light originates), its distance $u$ is always negative.

Important

Under this convention, the focal length of a concave mirror is negative (its focus $F$ sits in front of the mirror, on the same side as the object), and the focal length of a convex mirror is positive (its focus lies behind the mirror). This is one of the most frequently tested facts in CBSE board exams.

Watch out

A very common mistake is writing $f$ as positive for a concave mirror because "it converges light." Convergence tells you the type of mirror, not the sign — the sign comes purely from where the focus physically sits relative to the pole, under the convention above.

Where Does the Formula Come From?

For a concave mirror, parallel rays from a distant object converge at the focus, a point at (signed) distance $f$ from the mirror. The derivation uses similar triangles from a ray diagram.

›Proof

Consider an object of height $h_o$ in front of a concave mirror. Draw the ray parallel to the axis: it reflects through the focus $F$ . Draw the ray through the centre of curvature $C$ : it strikes the mirror normally and reflects straight back on itself. These two reflected rays cross to form the image, of height $h_i$ .

From similar triangles formed by the ray through $C$ :

$\frac{h_o}{h_i} = \frac{u - R}{R - v}$

where $R = 2f$ is the radius of curvature (with the same sign convention as $f$ ).

From similar triangles formed by the ray through $F$ :

$\frac{h_o}{h_i} = \frac{u - f}{f}$

Equating the two ratios and simplifying (using $R = 2f$ ) gives:

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

What the Equation Tells You

Rearranging for $v$ :

v = \frac{uf}{u - f}

Because $u$ is negative for a real object, and $v$ takes the sign the geometry dictates:

$v$ negative → the image forms in front of the mirror → real image (can be projected on a screen).
$v$ positive → the image forms behind the mirror → virtual image.

For a concave mirror ( $f$ negative), using the magnitude of the object distance $|u|$ measured from the pole:

$|u| > 2|f|$ → real, inverted, diminished image between $f$ and $2f$
$|u| = 2|f|$ → real, inverted, same-size image at $2f$
$|f| < |u| < 2|f|$ → real, inverted, magnified image beyond $2f$
$|u| = |f|$ → image at infinity …

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

The spherical mirror equation — also called the mirror formula — relates the object distance ( $u$ ), image distance ( $v$ ), and focal length ( $f$ ) of a spherical mirror. Let's build the reasoning step by step.

1. The Key Formula

For a spherical mirror (concave or convex):

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

Where:

$f$ = focal length (positive for concave, negative for convex)
$u$ = object distance from pole (always negative by sign convention)
$v$ = image distance from pole (sign depends on image location)

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Consider a concave mirror with:

Pole $P$
Centre of curvature $C$ (radius $R$ )
Focus $F$ (midpoint of $PC$ , so $f = R/2$ )

Take an object placed beyond $C$ . Draw two rays from the object's tip:

A ray parallel to the principal axis → reflects through $F$
A ray through $C$ → reflects back along itself

These rays meet at the image point.

Step 2: Use similar triangles

Let the object height be $h_o$ and image height be $h_i$ .

From the geometry of the ray through $C$ :

Triangle formed by object, $C$ , and axis is similar to triangle formed by image, $C$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - R}{R - v}

(Here $u$ and $v$ are distances from $P$ , with sign conventions applied later.)

Step 3: Use the parallel ray

From the ray parallel to the axis:

Triangle formed by object, $F$ , and axis is similar to triangle formed by image, $F$ , and axis.

This gives:

\frac{h_o}{h_i} = \frac{u - f}{f}

Step 4: Equate the two ratios

Since both ratios equal $h_o/h_i$ :

\frac{u - R}{R - v} = \frac{u - f}{f}

Step 5: Substitute $R = 2f$

For a spherical mirror, the focal length is half the radius of curvature:

R = 2f

Substitute:

\frac{u - 2f}{2f - v} = \frac{u - f}{f}

Step 6: Cross-multiply and simplify

Cross-multiply:

f(u - 2f) = (u - f)(2f - v)

Expand:

fu - 2f^2 = 2fu - uv - 2f^2 + fv

Cancel $-2f^2$ on both sides:

fu = 2fu - uv + fv

Bring all terms to one side:

0 = fu - uv + fv

Rearrange:

uv = fu + fv

Step 7: Divide by $uvf$

Divide both sides by $uvf$ :

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

This is the mirror formula.

3. Why the Sign Convention Matters

The derivation above used distances as positive magnitudes. In actual problem-solving, we use the Cartesian sign convention:

Distances measured against incident light are negative
Distances measured along incident light are positive

For a concave mirror:

$u$ is negative (object in front)
$f$ is negative (focus in front)
$v$ is negative for real images (in front) …

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A rainbow is intrinsically a full circle about the antisolar point; from an aeroplane there is no horizon to cut off its lower half, so a passenger can see the primary and secondary bows as complete concentric circles — option (b).

Why a rainbow is a circle

Sunlight entering a spherical water droplet is refracted on entry, reflected once (primary bow) or twice (secondary bow) inside, and refracted again on leaving. Because of dispersion each colour emerges at a slightly different angle, but every drop that sends light to your eye lies at a fixed angle from the line joining the Sun to your eye (the antisolar direction): about $42^\circ$ for the primary and about $51^\circ$ for the secondary. All such drops therefore lie on a cone whose axis is that antisolar line — geometrically a complete circle in the sky.

What each observer sees …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Q. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other side of the lens, at its focus a convex mirror is placed such that final image formed coincides with the object. The focal length of convex mirror is (A) 20 cm (B) 10 cm (C) 15 cm (D) 30 cm$
›Reveal solutionSolution
This tests combining the lens formula with the retro-reflection (normal-incidence) condition at a mirror. Answer: 10 cm.

Concept and Intuition

If the final image coincides with the original object, light must retrace its exact path after reflecting off the mirror. A ray retraces its path off a spherical mirror only if it strikes the mirror normally, i.e. it is travelling along a radius of the mirror — which means the point the rays are converging toward (before the mirror intervenes) must be the mirror's centre of curvature $C$ . So the strategy is: (1) find where the lens alone would have focused the light, (2) find how far that point lies from the mirror, and (3) equate that distance to $R$ , giving $f_{mirror}=R/2$ .

Step-by-Step Solution

Lens formula (sign convention: distances measured from the lens, light travelling left to right): $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ , with $u=-15$ cm, $f=+10$ cm (convex lens).

$\dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{-15}\cdot(-1)$ ... more carefully: $\dfrac1v = \dfrac1f + \dfrac1u = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3-2}{30}=\dfrac{1}{30}$ , so $v=30$ cm.

So without the mirror, the lens would converge light to a real image 30 cm from the lens, on the far side.

The convex mirror is placed at the lens's focus on that side, i.e. at 10 cm from the lens. The converging beam is intercepted 10 cm before it would have converged, so relative to the mirror the rays are aiming at a point $30 - 10 = 20$ cm behind the mirror. This is a virtual object for the mirror. …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$5\ ms^{-1}$
›Reveal solutionSolution
Tests longitudinal magnification for a moving object in front of a mirror: the image speed is $(v/u)^2$ times the object speed. Answer: $45\ m/s$ .

Concept and Intuition

For a mirror, the image and object distances are related by the mirror formula $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ . As the object moves slowly along the axis, the image also moves — but not at the same speed, because the relationship between $v$ and $u$ is nonlinear. Differentiating the mirror equation with respect to time shows that the instantaneous ratio of image speed to object speed equals the square of the ratio $v/u$ (this ratio is also, incidentally, the linear magnification's magnitude). This squared dependence is why images near a focal point can appear to move dramatically faster or slower than the object itself.

Step-by-Step Solution

Find the image distance using the mirror formula (magnitude form, real object and real image, concave mirror): $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ , or equivalently $v=\dfrac{uf}{u-f}$ .

Substitute $u=20\ cm$ , $f=15\ cm$ : $v = \dfrac{20\times15}{20-15} = \dfrac{300}{5}=60\ cm$ .

Differentiate $\dfrac1v+\dfrac1u=\dfrac1f$ with respect to time (treating $f$ as constant): …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$24\ mm$
›Reveal solutionSolution
This is the classic NCERT 'dentist's mirror' problem: a concave mirror used with the object inside the focal length gives a magnified, virtual image. Answer: magnification = 3.

Concept and Intuition

A dentist uses a concave mirror held close to the tooth (object distance less than the focal length) so that the mirror produces a virtual, erect, and magnified image — letting the dentist see an enlarged view of the cavity. This is a standard application of the mirror formula in the regime $u < f$ .

Step-by-Step Solution

Using the Cartesian sign convention (distances measured from the pole, incident light direction positive, real objects/distances to the left are negative): for a concave mirror, $f = -24$ mm; the object is at $u = -16$ mm.

Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ .

$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-24} - \dfrac{1}{-16} = -\dfrac{1}{24} + \dfrac{1}{16}$ .

Common denominator 48: $-\dfrac{2}{48} + \dfrac{3}{48} = \dfrac{1}{48}$ , so $v = 48$ mm.

Since $v$ is positive (opposite sign convention region from a real image), this indicates a virtual image formed behind the mirror. …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$x$
›Reveal solutionSolution
Setting up the real-image magnification relation $m=f/(u-f)$ for both given object positions and solving the two equations simultaneously gives $f=30$ cm. Answer: (D).

Concept and Intuition

For a concave mirror forming a real (inverted) image, the magnification and object distance are linked through the focal length by $m=\dfrac{f}{u-f}$ (using magnitudes, with $u>f$ so the object sits beyond the focus). Two different object positions with two different (given) magnifications give two independent equations in the two unknowns $x$ (the first object distance) and $f$ (focal length) — solving them together pins down both.

Step-by-Step Solution

First position, $u=x$ , $m=3$ : $3=\dfrac{f}{x-f} \Rightarrow 3(x-f)=f \Rightarrow 3x=4f \Rightarrow f=\dfrac{3x}{4}$ .

Second position, $u=x+5$ , $m=2$ : $2=\dfrac{f}{(x+5)-f} \Rightarrow 2[(x+5)-f]=f \Rightarrow 2x+10=3f \Rightarrow f=\dfrac{2x+10}{3}$ .

Equate the two expressions for $f$ : $\dfrac{3x}{4}=\dfrac{2x+10}{3} \Rightarrow 9x = 8x+40 \Rightarrow x=40$ . …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$5\ \mathrm{cm}$
›Reveal solutionSolution
For a convex mirror, the image of an extended object along the principal axis is found by locating the images of its endpoints using the mirror formula, then taking the difference of their image distances. Here, the image length is $\frac{10}{21}$ cm, so the correct option is (B).

We have a convex mirror with radius of curvature $R = 20\ \text{cm}$ , so its focal length is $f = \frac{R}{2} = 10\ \text{cm}$ . For a convex mirror, the focal length is taken as positive in the standard sign convention (since the focus lies behind the mirror). The rod AB lies along the principal axis, with end B at 20 cm from the pole (so $u_B = -20\ \text{cm}$ using the Cartesian sign convention: distances measured from the pole, positive in the direction of incident light, which is toward the mirror; here the object is on the left, so object distances are negative). End A is 5 cm farther from the mirror than B, so $u_A = -25\ \text{cm}$ .

The mirror formula is:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
where $v$ is the image distance (positive for virtual images behind a convex mirror).

Image of end B For $u_B = -20\ \text{cm}$ and $f = +10\ \text{cm}$ :

$\frac{1}{v_B} + \frac{1}{-20} = \frac{1}{10}$ $\frac{1}{v_B} = \frac{1}{10} + \frac{1}{20} = \frac{2+1}{20} = \frac{3}{20}$ $v_B = \frac{20}{3}\ \text{cm}$
This is positive, so the image of B is virtual, behind the mirror, at $\frac{20}{3}$ cm from the pole.

Image of end A For $u_A = -25\ \text{cm}$ :

$\frac{1}{v_A} + \frac{1}{-25} = \frac{1}{10}$ $\frac{1}{v_A} = \frac{1}{10} + \frac{1}{25} = \frac{5+2}{50} = \frac{7}{50}$ $v_A = \frac{50}{7}\ \text{cm}$
Again positive, so the image of A is also virtual, behind the mirror.

Length of the image Since both images lie on the same side of the mirror (behind it), the image of the rod is the segment between these two image points. The length is the absolute difference of their distances from the pole:
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\dfrac{f}{n^2} - nf$
›Reveal solutionSolution
Combining the mirror formula with the magnification formula for a convex mirror and solving for the object distance in terms of $f$ and $n$ gives $u=f-f/n$ .

Concept and Intuition

For mirrors (New Cartesian sign convention: distances against the incident-light direction are negative; for a convex mirror $f$ is positive; a real object always has $u$ negative), the governing relations are:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}, \qquad m = -\frac{v}{u}$
A convex mirror always forms a virtual, erect, diminished image of a real object, so its magnification is positive and less than 1 — exactly consistent with a size ratio $n<1$ . The goal is to eliminate $v$ between the two equations to express $u$ purely in terms of $f$ and $n$ .

Step-by-Step Solution

From the mirror formula: $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{u} = \dfrac{u-f}{fu}$ , so $v = \dfrac{fu}{u-f}$ .

Substitute into $m=-v/u$ : $m = -\dfrac{1}{u}\cdot\dfrac{fu}{u-f} = \dfrac{-f}{u-f} = \dfrac{f}{f-u}$ .

Set $m=n$ (given the image is $n$ times the object size): $n = \dfrac{f}{f-u}$ .

Solve for $u$ : $f-u = \dfrac{f}{n} \Rightarrow u = f - \dfrac{f}{n}$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$Q. An object is placed at a distance of 18 cm in front of a mirror. If the image is formed at a distance of 4 cm on the other side, then focal length, nature of the mirror and nature of image are respectively (A) 3.14 cm, concave mirror and real image (B) 3.14 cm, convex mirror and real image (C) 5.14 cm, convex mirror and virtual image (D) 5.14 cm, concave mirror and virtual image$
›Reveal solutionSolution
Applying the mirror formula with the object real ( $u=-18\,\text{cm}$ ) and the image virtual, behind the mirror ( $v=+4\,\text{cm}$ ) gives $f\approx+5.14\,\text{cm}$ — a convex mirror forming a virtual image.

Concept and Intuition

A real image can only form in front of a mirror (same side as the object); anything found behind the mirror must be virtual. Only convex mirrors (and concave mirrors with the object between pole and focus) can produce a virtual image; here we let the sign of $f$ that comes out of the calculation tell us which mirror type is actually consistent with the data.

Step-by-Step Solution

Sign convention: distances of real objects/images (in front, same side as incident light) are negative; distances of virtual images (behind the mirror) are positive.

Object: real, in front, so $u = -18\ \text{cm}$ .

Image: stated to form "on the other side," i.e., behind the mirror $\Rightarrow$ virtual, so $v = +4\ \text{cm}$ .

Mirror formula: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ .

$\dfrac{1}{4}+\dfrac{1}{-18} = \dfrac{1}{4}-\dfrac{1}{18} = \dfrac{9}{36}-\dfrac{2}{36}=\dfrac{7}{36}$

$f = \dfrac{36}{7} \approx 5.14\ \text{cm}$ , and it's positive. …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$Q. The relation between focal length$
(f) and radius of curvature (R) of a spherical mirror is (A) $R = \dfrac{f}{2}$ (B) $f = \dfrac{R}{2}$ (C) $R = f$ (D) $R = \dfrac{f}{3}$
›Reveal solutionSolution
The standard paraxial-ray relation for spherical mirrors is $f=R/2$ . Answer: (B).

Concept and Intuition

For rays close to the principal axis (paraxial approximation), a geometric analysis of reflection at a spherical mirror surface shows that the focal point lies exactly halfway between the pole and the centre of curvature.

Step-by-Step Solution

Consider a ray parallel to the principal axis striking the mirror close to the pole and reflecting through the focus.

Using the geometry of the mirror (radius $R$ , centre of curvature $C$ ) and the law of reflection, for small angles the focal length works out to $f = \dfrac{R}{2}$ . …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$30\ \text{cm}$
›Reveal solutionSolution
Using the convex-mirror magnification relation $m=\dfrac{f}{x+f}$ with $m=1/4$ and $f=30$ cm gives an object distance of 90 cm.

Concept and Intuition

A convex mirror always forms a virtual, erect, diminished image, so its magnification $m$ is positive and less than 1. Combining the mirror formula with the magnification definition gives a clean relation between object distance $x$ , focal length $f$ , and $m$ .

Step-by-Step Solution

Let object distance (measured as a positive magnitude) be $x$ ; using the Cartesian sign convention, $u=-x$ , and for a convex mirror $f=+30$ cm.

Mirror formula: $\dfrac1v+\dfrac1u=\dfrac1f \Rightarrow \dfrac1v = \dfrac1f-\dfrac1u = \dfrac1f+\dfrac1x = \dfrac{x+f}{fx} \Rightarrow v=\dfrac{fx}{x+f}$ .

Magnification: $m=-\dfrac{v}{u} = -\dfrac{fx/(x+f)}{-x} = \dfrac{f}{x+f}$ .

Given $m=\dfrac14$ : $\dfrac{f}{x+f}=\dfrac14 \Rightarrow 4f=x+f \Rightarrow x=3f$ . …
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Q.A passenger in an aeroplane shall

Concept understanding — Spherical Mirror Equation

The Spherical Mirror Equation: From Intuition to Formula

The Core Idea

The Sign Convention (New Cartesian Sign Convention)

Where Does the Formula Come From?

What the Equation Tells You

Why this formula?

Spherical Mirror Equation: Why the Formula Holds

1. The Key Formula

2. Why This Formula Holds — The Derivation

Step 1: Start with a ray diagram

Step 2: Use similar triangles

Step 3: Use the parallel ray

Step 4: Equate the two ratios

Step 5: Substitute $R = 2f$

Step 6: Cross-multiply and simplify

Step 7: Divide by $uvf$

3. Why the Sign Convention Matters

Why a rainbow is a circle

What each observer sees …

Method: Reasoning About Atmospheric-Optics Geometry from an Elevated Viewpoint

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.A passenger in an aeroplane shall

Why a rainbow is a circle

What each observer sees …

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