Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is

Q: Why is the focal length taken as positive?

For a convex mirror, the center of curvature and the principal focus lie behind the mirror. According to the Cartesian sign convention, distances measured in the direction of incident light are positive. Hence, the focal length is positive.

Q: Why did we get negative values for the object positions?

The negative sign simply indicates that the object is placed in front of the mirror, which is opposite to the direction of the incident light. This perfectly aligns with our standard sign convention.

Q: Is the speed of the image constant?

No! The speed of the image is given by $v_{image} = -m^2 v_{object}$. Since the magnification $m$ changes as the object moves, the speed of the image also changes continuously.

Q: Why do we multiply by 18/5 to convert m/s to km/h?

To convert meters to kilometers, we divide by 1000. To convert seconds to hours, we divide by 3600. So, $1 \text{ m/s} = \frac{1/1000 \text{ km}}{1/3600 \text{ h}} = \frac{3600}{1000} \text{ km/h} = \frac{18}{5} \text{ km/h}$.

Q: What would happen if this was a concave mirror?

If it were a concave mirror, the focal length would be negative. The image could be real or virtual depending on the object's position, which would completely change the values of $v$ and $u$!

Question

Image of an object approaching a convex mirror of radius of curvature $20	ext{ m}$ along its optical axis is observed to move from $\frac{25}{3}	ext{ m}$ to $\frac{50}{7}	ext{ m}$ in $30	ext{ s}$. What is the speed of the object in $	ext{km h}^{-1}$?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a large convex mirror, like the ones you see at sharp turns on roads.

The mirror has a radius of curvature of $20	ext{ m}$, which immediately tells us that its focal length is $f = +10	ext{ m}$.

We are given the initial and final positions of the image formed by this mirror, and we need to find out how fast the actual object is moving.

### The Master Equation

To find the speed of the object, we first need to pinpoint its exact locations at the start and end of the $30	ext{ s}$ time interval.

This is where our trusty mirror formula comes into play.

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

By rearranging this equation, we can easily solve for the object distance $u$.

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

### Finding the Initial Position

Let's plug in the values for the initial state.

We know the focal length $f = 10	ext{ m}$ and the initial image position $v_1 = \frac{25}{3}	ext{ m}$.

$$ \frac{1}{u_1} = \frac{1}{10} - \frac{3}{25} $$

Taking the LCM as $50$, the numerator becomes $5 - 6$.

$$ \frac{1}{u_1} = -\frac{1}{50} $$

This tells us that the initial position of the object was $u_1 = -50	ext{ m}$.

The negative sign simply means the object is placed in front of the mirror, perfectly following our sign convention.

### Finding the Final Position

Now, let's repeat the process for the final state.

The image has moved to a new position, $v_2 = \frac{50}{7}	ext{ m}$.

$$ \frac{1}{u_2} = \frac{1}{10} - \frac{7}{50} $$

Again, using $50$ as the LCM, the numerator becomes $5 - 7$.

$$ \frac{1}{u_2} = -\frac{2}{50} = -\frac{1}{25} $$

So, the final position of the object is $u_2 = -25	ext{ m}$.

### Final Calculation

The object moved from $-50	ext{ m}$ to $-25	ext{ m}$.

This means the total distance covered by the object is $\Delta u = 25	ext{ m}$.

$$ \Delta u = |-25 - (-50)| = 25	ext{ m} $$

Since this distance was covered in $30	ext{ s}$, we can calculate the speed in meters per second.

$$ v_{obj} = \frac{25}{30} = \frac{5}{6}	ext{ m s}^{-1} $$

Finally, to convert this speed into kilometers per hour, we multiply by $\frac{18}{5}$.

$$ v_{obj} = \frac{5}{6} 	imes \frac{18}{5} = 3	ext{ km h}^{-1} $$

**The speed of the object is exactly $3	ext{ km h}^{-1}$.**

Animated Solution for Physics - Optics: Image of an object approaching a convex mirror of radius of curvature $20 m$ along its optical axis is observed to move from $\frac{25}{3} m$ to $\frac{50}{7} m$ in $30 s$ . What is the speed of the object in $km h^{- 1}$ ?

Enter Numerical Value:

Visualized Solution

$f = \frac{R}{2}$

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

$\frac{1}{u _{1}} = \frac{1}{f} - \frac{1}{v _{1}}$

$u_{1} = - 50 m$

$\frac{1}{u _{2}} = \frac{1}{f} - \frac{1}{v _{2}}$

$u_{2} = - 25 m$

$v_{o bj} = \frac{Δ u}{Δ t}$

$v_{o bj} = 3 km h^{- 1}$

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Spherical Mirror

Analyzing the Setup

The Master Equation

Finding the Initial Position

Finding the Final Position

Final Calculation

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