Understanding Convex Mirror Magnification

The question asks us to compare the positions of two points, P and Q, based on the magnification produced by a convex mirror when an object is placed at these points. We are given that the magnification is $1/3$ when the object is at P, and $1/4$ when the object is at Q.

A convex mirror is a diverging mirror. It always forms a virtual, erect, and diminished image, regardless of the object's position. The image is always located behind the mirror, between the pole and the principal focus.

For a mirror, the linear magnification ($m$) is defined as the ratio of the image height ($h'$) to the object height ($h$), or equivalently, the negative ratio of the image distance ($v$) to the object distance ($u$).

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

Another useful formula relating magnification, focal length ($f$), and object distance ($u$) is:

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

For a convex mirror, the focal length $f$ is always positive (conventionally, since the principal focus is behind the mirror). The object distance $u$ is always negative (conventionally, since the object is in front of the mirror).

Let's analyze the formula $m = \frac{f}{f-u}$ for a convex mirror:

• $f$ is positive.
• $u$ is negative.
• Therefore, $f-u = f - (\text{negative value}) = f + (\text{positive value})$. This means $f-u$ is always positive and greater than $f$.
• The magnification $m = \frac{f}{f-u}$ will always be positive (since $f$ is positive and $f-u$ is positive). This confirms that the image formed by a convex mirror is always erect.
• Since $f-u > f$, the fraction $\frac{f}{f-u}$ will always be less than 1. This confirms that the image formed by a convex mirror is always diminished (magnification is less than 1). So, for a convex mirror, $0 < m < 1$.

Now, let's consider how the magnification $m$ changes as the object distance $|u|$ changes. Remember that $u$ is negative. As the object moves farther away from the mirror, the absolute value of $u$ ($|u|$) increases. Since $u$ is negative, this means $u$ becomes a larger negative number (e.g., from -10 cm to -20 cm). This makes the denominator $f-u$ larger.

$$ m = \frac{f}{\text{larger denominator}} $$

If the denominator $f-u$ increases while the numerator $f$ is constant, the value of the fraction $m$ decreases.

Conclusion: For a convex mirror, as the object moves farther away (i.e., $|u|$ increases, $u$ becomes more negative), the magnification $m$ decreases.

Comparing Magnifications at P and Q

We are given the magnification at P is $m_P = 1/3$ and the magnification at Q is $m_Q = 1/4$.

Comparing these values:

$$ \frac{1}{3} \text{ versus } \frac{1}{4} $$

Since $1/3 = 0.333...$ and $1/4 = 0.25$, we have $1/3 > 1/4$.

So, the magnification at P ($m_P$) is greater than the magnification at Q ($m_Q$).

$$ m_P > m_Q $$

Determining Object Distances $|u_P|$ and $|u_Q|$

We know that for a convex mirror, smaller magnification corresponds to a larger object distance from the mirror.

Since $m_Q < m_P$, the object distance at Q ($|u_Q|$) must be greater than the object distance at P ($|u_P|$).

$$ |u_Q| > |u_P| $$

This means point Q is farther from the convex mirror than point P.

Analyzing the Options

Let's examine the given options based on our conclusion:

• Option 1: P is far from the mirror as compared to Q. This is incorrect because we found Q is farther than P.
• Option 2: Q is far from the mirror as compared to P. This aligns with our conclusion that $|u_Q| > |u_P|$. This option is correct.
• Option 3: P and Q both are coinciding. This is incorrect because they produce different magnifications ($1/3$ and $1/4$). If they coincided, the object distance would be the same, resulting in the same magnification.
• Option 4: Convex mirror produces the magnification always greater than 1. This is incorrect. As we discussed, a convex mirror always produces a diminished image, meaning the magnification $m$ is always less than 1 ($0 < m < 1$).

Therefore, the correct statement is that Q is far from the mirror as compared to P.

Revision Table: Convex Mirror Properties & Magnification

Additional Information: Magnification Formula Derivation

The magnification formula $m = \frac{f}{f-u}$ can be derived from the mirror formula and the basic magnification definition.

Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Rearranging to find $v$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u-f}{fu}$

So, $v = \frac{fu}{u-f}$.

Basic magnification definition: $m = -\frac{v}{u}$

Substitute the expression for $v$ into the magnification formula:

$$ m = -\frac{\left(\frac{fu}{u-f}\right)}{u} = -\frac{fu}{u(u-f)} = -\frac{f}{u-f} = \frac{f}{-(u-f)} = \frac{f}{f-u} $$

This derivation shows how the magnification formula used is consistent with the fundamental mirror equation and magnification definition.

Understanding this relationship between magnification and object distance is key to solving problems involving convex mirrors.