When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distan

Q: Why is the speed formula $v_i = m^2 v_o$?

By differentiating the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with respect to time, we get $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$. Rearranging this gives $\frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt}$. Since $m = -\frac{v}{u}$, this becomes $\vec{v}_i = -m^2 \vec{v}_o$. The magnitude gives the speed relation $v_i = m^2 v_o$.

Q: Why do we use $m^2$ instead of just $m$?

Transverse magnification $m$ relates heights perpendicular to the principal axis ($h_i = m h_o$). Longitudinal magnification relates lengths along the principal axis. Because of the geometry of spherical mirrors, the longitudinal magnification is proportional to the square of the transverse magnification ($m^2$).

Q: Does the image move in the same direction as the object?

No, because of the negative sign in the velocity relation $\vec{v}_i = -m^2 \vec{v}_o$, the image always moves in the opposite direction to the object relative to the mirror. If the object moves left, the image moves right.

Q: What if the object was moving perpendicular to the principal axis?

If the object moves perpendicular to the principal axis, we would use the transverse magnification directly. The velocity of the image would be $\vec{v}_{i,\perp} = m \cdot \vec{v}_{o,\perp}$.

Q: Why are $u$ and $v$ taken as negative?

According to the Cartesian sign convention, the pole of the mirror is the origin. Distances measured in the direction of incident light are positive, and those opposite are negative. Since the object and real image are in front of the concave mirror, their positions are on the negative x-axis.

Question

When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of $9 	ext{ cms}^{-1}$, the speed (in $	ext{cms}^{-1}$) with which image moves at that instant is ........... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a concave mirror. We have an object placed $30 	ext{ cm}$ in front of it, and its image is formed at $10 	ext{ cm}$. The object is moving with a speed of $9 	ext{ cm/s}$. We need to find how fast the image is moving at that exact instant.

To visualize this, think of the principal axis as a track. The object is cruising along this track, and the image is responding to this movement. But they don't move at the same speed! The speed of the image depends on how much the mirror magnifies the object at that specific position.

### The Master Equation

To find the speed of the image, we use the concept of **longitudinal magnification**. While transverse magnification tells us how tall the image is compared to the object, longitudinal magnification tells us how long the image is along the principal axis.

For small movements, the speed of the image $v_i$ is related to the speed of the object $v_o$ by the square of the transverse magnification $m$:

$$v_i = v_o \cdot m^2$$

This is our master equation. It tells us that if we know the magnification and the object's speed, we can easily find the image's speed.

### Calculating Transverse Magnification

First, we need to find the transverse magnification, $m$. The formula for a spherical mirror is:

$$m = -\frac{v}{u}$$

Let's substitute the values of $v$ and $u$. According to the Cartesian sign convention, both distances are negative because they are measured in front of the mirror (opposite to the direction of incident light).

$$m = -\frac{-10}{-30}$$

Simplifying this, the negative signs cancel out, and we get:

$$m = -\frac{1}{3}$$

This negative sign indicates that the image is real and inverted.

### Final Calculation

Now let's bring back our speed equation. We substitute the speed of the object ($v_o = 9 	ext{ cm/s}$) and the magnification we just found.

$$v_i = 9 	imes \left(-\frac{1}{3}ight)^2$$

The square of $-\frac{1}{3}$ is $\frac{1}{9}$.

$$v_i = 9 	imes \frac{1}{9}$$

Multiplying $9$ by $\frac{1}{9}$ gives us exactly $1$.

$$v_i = 1 	ext{ cm/s}$$

So, the speed of the image is **$1 	ext{ cm/s}$**.

As a bonus insight, if we look at the velocity vectors, the actual relation is $\vec{v}_i = -m^2 \vec{v}_o$. Because of the minus sign, the image always moves in the opposite direction to the object along the principal axis. If the object moves towards the mirror, the image moves away!

Enter Numerical Value:

Visualized Solution

Visual Anchor

Logic Bridge

Raw Setup

Atomic Compute

Atomic Compute

Atomic Compute

Final Answer

The Way Forward

The Sigma Insight: Spherical Mirror

Analyzing the Setup

The Master Equation

Calculating Transverse Magnification

Final Calculation

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