Q.Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R = 2\ \text{m}$. If the jogger is running at a speed of $5\ \text{m s}^{-1}$, how fast the image of the jogger appear to move when the jogger is

The side mirror is convex, $R=2\ \text{m}\Rightarrow f=+1\ \text{m}$. With object distance $u=-d$ and the jogger approaching, $d$ decreases so $u$ increases: $\frac{du}{dt}=+5\ \text{m s}^{-1}$.

Differentiating $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $\left|\frac{dv}{dt}\right|=\left(\frac{f}{u-f}\right)^2\left|\frac{du}{dt}\right|=\frac{5}{(d+1)^2}\ \text{m s}^{-1}$.

• (a) $d=39\ \text{m}$: $\frac{5}{1600}=0.003125\ \text{m s}^{-1}$
• (b) $d=29\ \text{m}$: $\frac{5}{900}\approx0.00556\ \text{m s}^{-1}$ …

Setting up

A side-view mirror is convex, $R=2\ \text{m}$, so $f=R/2=+1\ \text{m}$ (positive by the sign convention, since a convex mirror's focus lies behind it).

Using the Cartesian sign convention, an object a distance $d$ in front of the mirror has $u=-d$. As the jogger approaches, $d$ decreases, so $u$ (which is negative) is becoming less negative — i.e. $u$ is increasing with time:

(The jogger's own speed is $5\ \text{m s}^{-1}$ toward the mirror, which is exactly a rate of increase of $u$.)

Differentiating the mirror equation

Start from

Differentiate with respect to time (with $f$ constant):

Since $v=\frac{fu}{u-f}$, we have $\frac{v}{u}=\frac{f}{u-f}$, so

Evaluating at each distance

With $f=1\ \text{m}$ and $u=-d$, $u-f=-(d+1)$, so $\left(\frac{f}{u-f}\right)^2 = \frac{1}{(d+1)^2}$, and the image speed is $\frac{5}{(d+1)^2}\ \text{m s}^{-1}$.

Method: Differentiation of the Spherical Mirror Equation

We use the mirror equation and differentiate it with respect to time to relate object velocity and image velocity.

Step 1: Identify mirror type and sign conventions

• Side view mirror of a car is convex.
• Radius of curvature $R = +2\ \text{m}$ (convex mirror → $R$ positive).
• Focal length:

• Sign convention:
  • Object distance $u$ is negative (object in front of mirror).
  • Image distance $v$ is positive (virtual image behind mirror).

Step 2: Write the mirror equation

With $u$ negative, let $u = -x$ where $x$ is the positive distance of jogger from mirror.

Step 3: Solve for $v$ in terms of $x$

Here $f = 1\ \text{m}$, so:

Step 4: Differentiate with respect to time

We need $\frac{dv}{dt}$ (speed of image). Given $\frac{dx}{dt} = -5\ \text{m/s}$ (jogger approaching → $x$ decreasing).

Using quotient rule:

Substitute $\frac{dx}{dt} = -5$:

The negative sign means the image moves towards the mirror (virtual image gets smaller). …

Common Mistakes with the Spherical Mirror Equation (and How to Avoid Them)

This problem is a classic application of the mirror formula combined with differentiation (rate of change). Here are the most frequent errors students make:

1. Confusing Sign Convention for Convex Mirrors

The Mistake:

Students treat the side-view mirror as a concave mirror or forget that for a convex mirror, the focal length ($f$) is negative.

Why it happens:

The formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ is the same for both, but the sign of $f$ changes. A side-view mirror is convex (diverging).

How to avoid:

• Always identify the mirror type first.
• For a convex mirror:
  • $R$ is positive in magnitude, but $f = -\frac{R}{2}$ (by Cartesian sign convention).
  • Here, $R = 2\ \text{m}$, so $f = -1\ \text{m}$.
• Write the sign explicitly before substituting.

2. Forgetting that Object Distance ($u$) is Negative

The Mistake:

Plugging $u = +39\ \text{m}$ directly into the formula.

Why it happens:

In the Cartesian sign convention, distances measured against the incident light direction are negative. The object is in front of the mirror, so $u$ is negative.

How to avoid:

• Always set $u = -d$ where $d$ is the actual distance.
• For this problem: $u = -39,\ -29,\ -19,\ -9\ \text{m}$.

3. Using the Wrong Formula for Image Velocity

The Mistake:

Directly differentiating $v = \frac{uf}{u-f}$ without considering the sign of $f$ or using $v = \frac{f u}{u - f}$ incorrectly.

Why it happens:

Students often memorise $v = \frac{uf}{u-f}$ but forget that this form assumes $f$ is positive (concave). For convex mirrors, the correct derived relation is:

but with $f$ negative, the denominator becomes $u - (-|f|) = u + |f|$, which changes the sign of the result.

How to avoid:

• Derive from the mirror formula each time:

Then solve for $v$ after substituting signed values.

• Then differentiate:

(This formula is sign-sensitive — use it carefully.)

4. Ignoring the Sign of $\frac{du}{dt}$

The Mistake:

Taking $\frac{du}{dt} = +5\ \text{m/s}$ when the jogger is approaching.

Why it happens:

If the jogger is moving toward the mirror, the distance $|u|$ decreases, but $u$ itself (which is negative) becomes less negative — so $\frac{du}{dt}$ is positive.

How to avoid:

• Define $u = -x$, where $x$ is the positive distance.
• Then $\frac{du}{dt} = -\frac{dx}{dt}$.
• If the jogger approaches, $\frac{dx}{dt} = -5\ \text{m/s}$ (since $x$ decreases), so $\frac{du}{dt} = -(-5) = +5\ \text{m/s}$.
• Key: $\frac{du}{dt} = +5\ \text{m/s}$.

5. Forgetting That Image Velocity is Also Negative

The Mistake:

Reporting the speed as a positive number without realising the image moves toward the mirror (opposite direction to the object's approach).

Why it happens:

Students compute $\frac{dv}{dt}$ and get a negative value, then panic or drop the sign.

How to avoid: …