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.

COMEDK 2026Set 2026-M1 markMCQ
$Q. An object is placed 25 cm in front of a fully silvered concave mirror of focal length 15 cm . A plane mirror is placed 35 cm behind the concave mirror on the side opposite to the object. The final image after reflection first from the concave mirror and then from the plane mirror is formed at: (A) 37.5 cm in front of the plane mirror (B) 72.5 cm behind the concave mirror (C) 107.5 cm behind the concave mirror (D) 37.5 cm in front of the concave mirror$
›Reveal solutionSolution
The key idea is to treat the fully silvered concave mirror as a mirror (not a lens) and then use the plane mirror to fold the path. After two reflections, the final image lies 107.5 cm behind the concave mirror, which corresponds to option (C).

Concept & Intuition

A fully silvered concave mirror is just a concave mirror — it reflects light. The plane mirror is placed behind it, so after the first reflection from the concave mirror, the light travels toward the plane mirror, reflects again, and the final image location is found by applying the mirror formula twice, carefully tracking sign conventions. The trick: the plane mirror simply creates a virtual image at the same distance behind it as the object is in front of it. We must keep distances measured from the concave mirror consistent.

Step-by-step solution

First reflection from the concave mirror

Object distance from concave mirror: $u_1 = -25$ cm (negative because object is in front, per Cartesian sign convention).

Focal length: $f = -15$ cm (concave mirror, negative focal length).

Mirror formula: $\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f}$ .

$\frac{1}{v_1} + \frac{1}{-25} = \frac{1}{-15}$ $\frac{1}{v_1} = -\frac{1}{15} + \frac{1}{25} = \frac{-5 + 3}{75} = -\frac{2}{75}$
So $v_1 = -37.5$ cm.

The negative sign means the image is in front of the concave mirror (real image), 37.5 cm from it.

Position of this image relative to the plane mirror

The plane mirror is 35 cm behind the concave mirror. So the distance from the concave mirror to the plane mirror is $+35$ cm (taking direction from concave toward plane as positive).

The first image is at $-37.5$ cm from the concave mirror (i.e., 37.5 cm in front).

Distance from this image to the plane mirror = $35 - (-37.5) = 72.5$ cm.

So the image is 72.5 cm in front of the plane mirror.

Second reflection from the plane mirror …
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KCET 2026Set C21 markMCQ
$Q. The direction of a ray of light incident on a concave mirror is shown by PQ, while direction in which the ray would travel after reflection is shown by four rays marked as A, B, C and D as shown in the figure. Which of the four rays correctly shows the direction of the reflected ray? PQ is a paraxial ray. (A) D (B) C (C) B (D) A$
›Reveal solutionSolution

A basic property of a concave (converging) mirror: any paraxial ray travelling parallel to the principal axis, after reflection, passes through the mirror's principal focus F.

Step 1 — Recall the reflection rule for a ray parallel to the axis

For a concave mirror, a ray parallel to the principal axis strikes the mirror and is reflected such that the reflected ray passes through the focus F, which lies midway between the pole and the centre of curvature C on the principal axis. This follows directly from the law of reflection applied to the mirror's curved surface at the point of incidence Q.

Step 2 — Identify the correct ray among the four options …
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KCET 2025Set D-41 markMCQ
$n$
›Reveal solutionSolution

Put $i = 2r$ into Snell's law, expand $\sin 2r$ with the double-angle identity, cancel $\sin r$ , and solve for $r$ .

Step 1 — Write Snell's law for the vacuum $\to$ medium refraction.

Light goes from vacuum ( $n_1 = 1$ ) into a medium of refractive index $n$ :
$n_1 \sin i = n_2 \sin r \quad\Longrightarrow\quad \sin i = n \sin r$
Step 2 — Impose the given condition.

We are told the angle of incidence is twice the angle of refraction:
$i = 2r$
Substituting:
$\sin(2r) = n \sin r$
Step 3 — Use the double-angle identity.
$\sin 2r = 2\sin r\cos r$
so
$2\sin r\cos r = n\sin r$
Step 4 — Cancel $\sin r$ .

Since the ray actually refracts, $r \neq 0$ , so $\sin r \neq 0$ and we may divide both sides by it:
$2\cos r = n \quad\Longrightarrow\quad \cos r = \frac{n}{2}$ $r = \cos^{-1}\!\left(\frac{n}{2}\right)$
Step 5 — Return to the angle of incidence.

The question asks for $i$ , not $r$ :

$i = 2r = 2\cos^{-1}\!\left(\frac{n}{2}\right)$ …
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KCET 2025Set D-41 markMCQ
$P$
›Reveal solutionSolution

A cut along the principal axis does not change the power ( $L_1 = P$ ); a cut perpendicular to it halves the power ( $L_2 = L_3 = P/2$ ), so the incorrect statement is the one claiming $L_1$ has power $P/2$ .

Step 1 — The lens-maker's formula is the concept.
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right), \qquad P = \frac{1}{f}$
The power depends only on the refractive index and the two radii of curvature — not on the aperture (the size/height of the lens).

Step 2 — Cut ALONG the principal axis.

This cut is made in a plane containing the principal axis, so each half still has both curved surfaces with the same $R_1$ and $R_2$ ; only the aperture is halved. Therefore
$P_{half} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = P$
The piece $L_1$ (the half that is not cut again) has power $P$ . (Only its light-gathering area, and hence image brightness, is halved.)

Step 3 — Cut PERPENDICULAR to the principal axis.

The second half is now sliced through the middle perpendicular to the axis. Each resulting piece is plano-convex: one surface keeps its curvature, the other becomes flat ( $R \to \infty$ ).

For the original biconvex lens with $R_1 = +R$ , $R_2 = -R$ : …
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COMEDK 2025Set 2025-M1 markMCQ
$m_1$
›Reveal solutionSolution
The key is to relate the change in image distance to the two magnifications using the lens formula. The focal length turns out to be $f = \frac{x}{m_2 - m_1}$ , so the correct option is (C).

Concept and intuition

When a convex lens forms a real image, both object distance $u$ and image distance $v$ are positive (using the real-is-positive sign convention). The magnification for a real image is $m = v/u$ . If we move the lens toward the object, $u$ decreases, so $v$ must increase to keep the lens equation $1/f = 1/u + 1/v$ satisfied. The problem tells us that the image distance increases by $x$ , and gives two magnifications $m_1$ and $m_2$ . The trick is to express $u$ and $v$ in terms of $m$ and $f$ , then use the change in $v$ to solve for $f$ .

Step-by-step solution

Relate magnification to object and image distances For a real image formed by a convex lens,

$m = \frac{v}{u}.$
So we can write $v = m u$ .

Use the lens formula The lens equation is

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}.$
Substitute $v = m u$ :
$\frac{1}{f} = \frac{1}{u} + \frac{1}{m u} = \frac{1}{u}\left(1 + \frac{1}{m}\right) = \frac{m+1}{m u}.$
Hence
$u = \frac{m+1}{m} f.$
Then the image distance is
$v = m u = m \cdot \frac{m+1}{m} f = (m+1)f.$

Apply to the two positions For the first position: $v_1 = (m_1 + 1)f.$ …
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COMEDK 2024Set 2024-M1 markMCQ
$10 \mathrm{~cm}$
›Reveal solutionSolution
The focal length of a convex mirror depends only on its geometry (radius of curvature), not on the wavelength of light. Therefore, changing the wavelength does not change the focal length. The answer is 10 cm.

Concept and Intuition

The key idea here is to recall what determines the focal length of a mirror. For any spherical mirror — concave or convex — the focal length $f$ is given by $f = \frac{R}{2}$ , where $R$ is the radius of curvature of the mirror. This radius is a purely geometric property: it is the radius of the sphere from which the mirror’s surface is cut. It does not depend on the color, wavelength, or frequency of the light used.

A common confusion arises because lenses do have wavelength-dependent focal lengths (chromatic aberration), but mirrors work by reflection, not refraction. Reflection obeys the law of reflection for all wavelengths equally — there is no dispersion. So, no matter what light you shine on a mirror, its focal length stays fixed.

Let’s walk through the reasoning step by step.

Recall the mirror formula for focal length For any spherical mirror (concave or convex), the focal length is

$f = \frac{R}{2}$
where $R$ is the radius of curvature. This is derived from geometry and the law of reflection, and it contains no term involving wavelength.

Identify what changes when wavelength is doubled

The problem states that the wavelength of incident light is doubled, but the mirror’s area (and therefore its shape and radius) is kept constant. Changing the wavelength does not alter the physical shape of the mirror — the mirror remains the same piece of glass with the same silvered surface.

Check for any possible effect of wavelength on reflection …
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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

Step 1 — Recall the reflection rule for a ray parallel to the axis

Step 2 — Identify the correct ray among the four options …

Q.A passenger in an aeroplane shall

Why a rainbow is a circle

What each observer sees …

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