Question

An object placed at a distance of +15 cm is slowly moved towards the pole of a convex mirror. The image will get

• A. shortened and real.
• B. enlarged and real.
• C. enlarged and virtual.
• D. diminished and virtual.

Hint:

While a convex mirror always produces a diminished image compared to the actual object size, moving the object closer to the mirror increases the image size relative to its previous state.
At the pole, the image size equals the object size (\(m = 1\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks about the change in characteristics of the image when an object is moved closer to the pole of a convex mirror.

Step 2: Key Formula or Approach:
The behavior can be analyzed using the mirror formula and magnification equation:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
\[ m = -\frac{v}{u} \]

Step 3: Detailed Explanation:

• For a convex mirror, the focal length (\(f\)) is always positive, and the object distance (\(u\)) is negative according to the sign convention.

• Rearranging the mirror formula to find image distance (\(v\)):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{|u|} \]

• Since both \(f\) and \(|u|\) are positive, \(v\) is always positive. This means the image is always formed behind the mirror, making it virtual and erect.

• As the object is moved closer to the pole, the magnitude of the object distance (\(|u|\)) decreases.

• As \(|u|\) decreases, the term \(\frac{1}{|u|}\) increases, which increases \(\frac{1}{v}\), thereby causing the image distance \(v\) to decrease (the image moves closer to the pole).

• The magnification is given by:
\[ m = \frac{f}{f - u} = \frac{f}{f + |u|} \]

• As \(|u|\) decreases, the denominator (\(f + |u|\)) decreases, which increases the value of magnification \(m\).

• Consequently, the image becomes larger (enlarged compared to its initial size) while remaining virtual and erect.


Step 4: Final Answer:
The image will get enlarged and virtual.