Question

An object is placed at a distance of $20\,\text{cm}$ in front of a concave mirror of focal length $15\,\text{cm}$. What is the nature and magnification ($m$) of the image produced by the mirror?}

• A. \( \text{Virtual, Erect and } m = +3 \)
• B. \( \text{Real, Inverted and } m = -3 \)
• C. \( \text{Real, Inverted and } m = -0.75 \)
• D. \( \text{Virtual, Erect and } m = +0.75 \)

Hint:

Using the shortcut equation \(m = \frac{f}{f-u}\) lets you calculate magnification in one step, bypassing the need to solve for the intermediate image distance variable \(v\) first. This saves valuable time during competitive exams.

Answer 1

The Correct Option is B

Concept: To find the location and properties of an image formed by a spherical mirror, we use the standard mirror formula along with the Cartesian sign convention system: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where \(f\) is the focal length, \(u\) is the object distance, and \(v\) is the image distance. The Cartesian sign convention rules state:
• All distances are measured starting from the pole (\(P\)) of the mirror as the origin.
• Distances measured in the direction of the incoming light rays are positive.
• Distances measured against the direction of incoming light rays are negative.
• Consequently, real objects placed in front of a mirror always have a negative distance (\(u < 0\)).
• The focal length of a concave mirror is always negative (\(f < 0\)), while the focal length of a convex mirror is always positive (\(f > 0\)). The lateral magnification factor (\(m\)) relates the height of the image to the height of the object, and can be calculated using these distances: \[ m = -\frac{v}{u} = \frac{f}{f - u} \] A negative magnification value (\(m < 0\)) means the image is real and inverted, while a positive value (\(m > 0\)) means it is virtual and erect.

Step 1: Apply sign conventions to the given values. From the problem description, we can identify and sign-resolve our key parameters:
• Mirror type: Concave $\rightarrow$ Negative focal length value: \(f = -15\,\text{cm}\)
• Object location: In front of the mirror $\rightarrow$ Negative object distance value: \(u = -20\,\text{cm}\)

Step 2: Calculate the magnification factor using the focal length shortcut formula. Let's plug these values directly into the magnification shortcut equation to save time: \[ m = \frac{f}{f - u} \] \[ m = \frac{-15}{-15 - (-20)} \] Simplify the signs in the denominator: \[ m = \frac{-15}{-15 + 20} \] \[ m = \frac{-15}{5} \] \[ m = -3 \]

Step 3: Interpret the physical characteristics from the magnification value. We can break down our calculated value of \(m = -3\) to find the image properties:
• The negative sign (-): Confirms that the image is real and inverted.
• The absolute value magnitude (\(|m| = 3 > 1\)): Confirms that the image is magnified to three times the size of the object. This matches standard optics expectations: since the object is placed at \(20\,\text{cm}\), it sits between the focus (\(F = 15\,\text{cm}\)) and the center of curvature (\(C = 2f = 30\,\text{cm}\)). Any object in this zone forms a real, inverted, and magnified image located beyond \(C\).