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.

KEAM 2026Set eng-2026-04194 marksMCQ
$f$
›Reveal solutionSolution

An object at distance $f$ before a convex mirror of focal length $f$ forms its image at $f/2$ behind the mirror.

Mirror equation: $\dfrac1v+\dfrac1u=\dfrac1f$ .

For a convex mirror $f=+f$ ; the real object is at $u=-f$ .

$\frac1v=\frac1f-\frac1u=\frac1f-\frac1{-f}=\frac1f+\frac1f=\frac2f.$ …
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KEAM 2026Set eng-2026-04214 marksMCQ
$4$
›Reveal solutionSolution

Mirror formula gives $v=-40\,$ cm, magnification magnitude $4$ .

Using the mirror sign convention, $u=-10\,cm$ , $f=-8\,cm$ : …
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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. The distance of an object placed in front of a concave mirror of radius of curvature 24 cm that gives its magnification as 3 is (A) 8 cm (B) 16 cm (C) 12 cm (D) 24 cm (E) 32 cm$
›Reveal solutionSolution

Radius $24$ cm $\Rightarrow f=12$ cm. A stated (positive) magnification of 3 means an erect, virtual, enlarged image, which occurs with the object inside the focus; solving the mirror relation gives the object 8 cm from the mirror.

Focal length $f=\dfrac{R}{2}=12$ cm; with the mirror sign convention $f=-12$ cm.

Using $m=\dfrac{f}{f-u}$ and $u=f\dfrac{m-1}{m}$ . Taking the magnification as $+3$ (an erect, virtual, magnified image, as a concave mirror gives when the object lies between the pole and focus): …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$-0.5$
›Reveal solutionSolution
The magnification is $-2$ .

Concept and Intuition

For mirrors, $m=-\dfrac{v}{u}$ . A real image formed on the same side as the object (in front of a concave mirror) is inverted and, being farther out, enlarged.

Step-by-Step Solution

Sign convention: $u=-10\,$ cm, $v=-20\,$ cm (both in front).

$m=-\dfrac{v}{u}=-\dfrac{-20}{-10}=-2$ . …
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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

Q.A passenger in an aeroplane shall

Why a rainbow is a circle

What each observer sees …

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