Question

An object is 12 cm in front of a convex mirror with a center/radius of a curvature of 6 cm. The characteristics of the image are___A. real, inverted at a distance of 12 cmB. virtual, upright at a distance of 2.4 cmC. real, upright at a distance of 4 cmD. virtual, inverted at a distance of 6 cm Must follow GRESA format.Given:Required:Equation(s) Needed:Solution:Final Answer:L-ocation:O-rientation:S-izeT-ype

Answer 1

Given data:

* The radius of curvature of the convex mirror is R = 6 cm.

* The distance of the object from the convex mirror is u = - 12 cm.

Required: The nature of the image formed.

Equations needed:

The mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

The magnification formula:

$m=-\frac{v}{u}=\frac{h_i}{h_o}$

Solution:

The focal length of the convex mirror is,

$\begin{gathered} f=\frac{R}{2} \\ f=\frac{6}{2} \\ f=3\text{ cm} \end{gathered}$

The distance of the image formed is,

$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}+\frac{1}{(-12)}=\frac{1}{3} \\ \frac{1}{v}-\frac{1}{12}=\frac{1}{3} \\ \frac{1}{v}=\frac{1}{3}+\frac{1}{12} \end{gathered}$

By simplifying,

$\begin{gathered} \frac{1}{v}=\frac{4+1}{12} \\ \frac{1}{v}=\frac{5}{12} \\ v=\frac{12}{5} \\ v=2.4\text{ cm} \end{gathered}$

The positive value of image distance indicated that the virtual image is formed.

By the magnification formula,

$\begin{gathered} m=\frac{-(2.4)}{-12} \\ m=0.2 \end{gathered}$

The positive value of magnification indicates that the upright image is formed.

The value of the magnification is less than 1, thus, a small size (diminished) image is formed.

Final Answer:

Location - 2.4 cm behind the convex mirror

Orientation - Upright

Size - Small size (or diminished)

Type - Virtual

Hence, option B is the correct answer.