DCE 116 (Heat, Geometric Optics and Radioactivity) Instruction: Answer Three (3) Questions

Given:
Object distance, $u = 12$ cm
Focal length of convex mirror, $f = -15$ cm (focal length is negative for convex mirrors)
Object height, $h_o = 4.5$ cm

The mirror formula is:

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

where $v$ is the image distance.

Rearrange the mirror formula to solve for $v$:

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

Substitute the given values:

$$\frac{1}{v} = \frac{1}{-15 \text{ cm}} - \frac{1}{12 \text{ cm}}$$

Find a common denominator, which is 60:

$$\frac{1}{v} = \frac{-4}{60 \text{ cm}} - \frac{5}{60 \text{ cm}}$$ $$\frac{1}{v} = \frac{-4 - 5}{60 \text{ cm}}$$ $$\frac{1}{v} = \frac{-9}{60 \text{ cm}}$$ $$v = \frac{60 \text{ cm}}{-9}$$ $$v = -\frac{20}{3} \text{ cm} \approx -6.67 \text{ cm}$$

The negative sign for $v$ indicates that the image is formed behind the mirror, meaning it is a virtual image.

The magnification formula is:

$$M = -\frac{v}{u}$$

Substitute the values of $v$ and $u$:

$$M = -\frac{-\frac{20}{3} \text{ cm}}{12 \text{ cm}}$$ $$M = \frac{20}{3 \times 12}$$ $$M = \frac{20}{36}$$ $$M = \frac{5}{9} \approx 0.56$$

The positive sign for $M$ indicates that the image is upright.

The magnification formula also relates image height ($h_i$) to object height ($h_o$):

$$M = \frac{h_i}{h_o}$$

Rearrange to solve for $h_i$:

$$h_i = M \times h_o$$

Substitute the values of $M$ and $h_o$:

$$h_i = \frac{5}{9} \times 4.5 \text{ cm}$$ $$h_i = \frac{5}{9} \times \frac{9}{2} \text{ cm}$$ $$h_i = \frac{5}{2} \text{ cm}$$ $$h_i = 2.5 \text{ cm}$$

• Location: The image is located $\frac{20}{3}$ cm or approximately 6.67 cm behind the mirror.
• Size: The image height is 2.5 cm.
• Nature: The image is virtual and upright.
• Magnification: The magnification is $\frac{5}{9}$ or approximately 0.56.

Final Answer:
The location of the image is $\boxed{\text{6.67 cm behind the mirror}}$.
The size of the image is $\boxed{\text{2.5 cm}}$.
The nature of the image is $\boxed{\text{virtual and upright}}$.
The magnification of the image is $\boxed{\text{0.56}}$.