Correct Answer: Option A

Explanation

A. Less than 20 cm

For a concave mirror, an erect image is virtual (behind the mirror) and forms when the object is placed between the pole and focus (\( u < f \)).

Magnification:
\(m = +4 \quad (\text{erect, so positive})\)

Using the mirror formula:
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)

Given \( f = -20 \, \text{cm} \) (concave), \( u \) is negative.

Since \( m = -\frac{v}{u} \):
\(v = -m u = -(+4)u = -4u\)

Substituting into the mirror formula:
\(\frac{1}{-4u} + \frac{1}{u} = \frac{1}{-20}\)

Combining the fractions:
\(\frac{3}{4u} = \frac{1}{-20}\)

Thus, solving for \( u \):
\(3 \cdot -20 = 4u \quad \Rightarrow \quad u = -15 \, \text{cm}\)

Object distance \( |u| = 15 \, \text{cm} < 20 \, \text{cm} \).

Thus, it is less than 20 cm.

OR

A concave mirror only gives an erect, magnified image when the object is placed between the mirror and its focal point. Since the focal length is 20 cm, the object must be closer than 20 cm to the mirror.

There is an explanation video available below.

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