Step 1: Understanding the Concept:
To solve problems involving spherical mirrors, we must adhere strictly to the Cartesian sign convention:
1. The pole of the mirror is treated as the origin \((0,0)\).
2. Distances measured in the direction of incident light are positive, while those measured against it are negative.
3. Consequently, for a real object placed in front of a mirror, the object distance \(u\) is always negative.
4. For a concave mirror, the focus is in front of the reflecting surface, so the focal length \(f\) is always negative.
The lateral magnification \(m\) tells us how much larger or smaller the image is compared to the object.
The sign of \(m\) is critical: a negative value indicates a real, inverted image, while a positive value indicates a virtual, erect image.
Step 2: Key Formula or Approach:
We can combine the mirror formula (\(1/v + 1/u = 1/f\)) and the magnification formula (\(m = -v/u\)) into a highly efficient shortcut formula:
\[ m = \frac{f}{f - u} \]
This allows us to calculate magnification directly from the given parameters without first solving for the image distance \(v\).
Step 3: Detailed Explanation:
Let's extract the known values from the problem and apply the sign convention:
- Mirror Type: Concave \(\rightarrow f = -15\) cm.
- Object Location: In front of the mirror \(\rightarrow u = -20\) cm.
Now, substitute these into the shortcut magnification formula:
\[ m = \frac{-15}{(-15) - (-20)} \]
\[ m = \frac{-15}{-15 + 20} \]
\[ m = \frac{-15}{5} = -3 \]
Interpretation of the result \(m = -3\):
1. Negative Sign (-): This confirms that the image is real and inverted.
2. Magnitude (3): Since \(|m|>1\), the image is magnified (specifically, 3 times the size of the object).
This result aligns with ray optics theory. The center of curvature \(C\) for this mirror is at \(2f = 30\) cm. The object is placed at \(20\) cm, which is between the focus (\(15\) cm) and the center of curvature (\(30\) cm).
Objects placed between \(F\) and \(C\) always produce a real, inverted, and enlarged image located beyond \(C\).
If we were to calculate the image distance \(v\):
\[ m = -\frac{v}{u} \implies -3 = -\frac{v}{-20} \implies v = -60 \text{ cm} \]
The image is indeed formed at \(60\) cm (beyond \(C\)) on the same side as the object.
Step 4: Final Answer:
The nature of the image is real and inverted, and the magnification is \(-3\).