A car is moving with at a constant speed of $60 \text{ km h}^{-1}$ on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of $5 \text{ km h}^{-1}$. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement(s) is/are correct? (a) The speed of the car in the rear is 65 km h–1. (b) In the side mirror the car in the rear would appear to approach with a speed of 5 km h–1 to the driver of the leading car. (c) In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases. (d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

A car is moving with at a constant speed of 60 km h⁻¹ on a straight road. Looking at the rear view mirror, the

NCERT Exemplar · Q10

60 \text{ km h}^{-1}

(a) The speed of the car in the rear is 65 km h–1.

(b) In the side mirror the car in the rear would appear to approach with a speed of 5 km h–1 to the driver of the leading car.

(c) In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.

(d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

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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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The rear car's true road speed is $65\ \text{km h}^{-1}$ (statement a). The driver's interior rear-view mirror is a plane mirror, so the apparent closing speed there stays a constant $5\ \text{km h}^{-1}$ - it does not decrease. The side mirror is convex, and as the gap closes, its magnification grows, so the apparent speed in the side mirror increases (statement d). The correct statements are (a) and (d).

Statement (a): the rear car's true speed

Speeds relative to the road simply add. The leading (driver's) car travels at $60\ \text{km h}^{-1}$ ; the car behind is closing the gap (approaching) at a relative speed of $5\ \text{km h}^{-1}$ . So its speed over the ground is

v_{\text{rear}} = 60 + 5 = 65\ \text{km h}^{-1}.

Statement (a) is correct.

Statement (b): apparent speed in the side mirror

The side mirror is described as a convex mirror (standard for side/rear-view mirrors, to give a wider field of view). A convex mirror's image position does not move in direct proportion to the object's true motion - the image position and the true position are related nonlinearly through the mirror equation, so the image does not simply mirror the true relative speed of $5\ \text{km h}^{-1}$ . Statement (b), that the car appears to approach at a constant $5\ \text{km h}^{-1}$ in the side mirror, is incorrect.

Statement (c): rear-view mirror behaviour as distance shrinks

The interior rear-view mirror is a plane mirror. A plane mirror's image position exactly mirrors the object's true position at all times - image distance behind the mirror equals object distance in front of it, always, regardless of how close the object gets. So the apparent speed in a plane mirror is always the true relative speed, constant at $5\ \text{km h}^{-1}$ - it does not decrease as the gap shrinks. Statement (c) is incorrect.

Statement (d): apparent speed in the side (convex) mirror as distance shrinks …

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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

Statement (a): the rear car's true speed

Statement (b): apparent speed in the side mirror

Statement (c): rear-view mirror behaviour as distance shrinks

Statement (d): apparent speed in the side (convex) mirror as distance shrinks …

Method: Distinguishing True Relative Speed from Apparent Speed in Plane vs. Curved Mirrors

Steps

Statement (a): the rear car's true speed

Statement (b): apparent speed in the side mirror

Statement (c): rear-view mirror behaviour as distance shrinks

Statement (d): apparent speed in the side (convex) mirror as distance shrinks …

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