Question

An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the position and nature of the image.

• A. \(20\) cm in front, real and inverted
• B. \(30\) cm in front, real and inverted
• C. \(30\) cm behind, virtual and erect
• D. \(15\) cm behind, virtual and erect

Hint:

Apply the mirror formula with the Cartesian sign convention, taking distances in front of the mirror as negative. Once v is found, calculate the magnification m = -v/u to decide whether the image is inverted and enlarged.

Answer 1

The Correct Option is B

Concept: The mirror formula is \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] where \(f\) = focal length, \(u\) = object distance, \(v\) = image distance. For a concave mirror, distances measured in front of the mirror are negative according to the Cartesian sign convention.

Step 1: Substitute the given values. \[ u=-15\text{ cm}, \quad f=-10\text{ cm} \] Using the mirror formula: \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] \[ \frac{1}{-10}=\frac{1}{v}+\frac{1}{-15} \]

Step 2: Solve for \(v\). \[ \frac{1}{v}=\frac{1}{-10}+\frac{1}{15} \] \[ \frac{1}{v}=-\frac{1}{30} \] \[ v=-30 \text{ cm} \]

Step 3: Interpret the result. The negative sign indicates that the image forms in front of the mirror. Hence, the image is real and inverted.

Answer 2

Concept:

• The mirror formula can be rearranged into a direct formula for image distance: $v = \dfrac{uf}{u-f}$, which avoids solving the reciprocal equation step by step.
• The nature of the image (real or virtual, inverted or erect) is confirmed using the linear magnification $m = -\dfrac{v}{u}$, not just by reading the sign of v.

Step 1: Note the given values with sign convention.
Object distance $u = -15\,cm$, focal length $f = -10\,cm$

Step 2: Find the image distance using the direct formula.
$v = \dfrac{uf}{u-f} = \dfrac{(-15)(-10)}{(-15)-(-10)} = \dfrac{150}{-5} = -30\,cm$

Step 3: Find the magnification to confirm the nature of the image.
$m = -\dfrac{v}{u} = -\dfrac{(-30)}{(-15)} = -2$
The negative sign of m shows the image is inverted, and $|m| = 2$ shows it is magnified. Since v is negative, the image forms in front of the mirror, so it is real.

Final Answer: The image forms 30 cm in front of the mirror, and is real and inverted.