Q8E - University Physics with Modern Physics

Question

An object is 18cm from the centre of a spherical silvered-glass Christmas tree ornament 6cm in diameter. What are the position and magnification of its image?

Step-by-Step Solution

Answer

The position is $1.36cm$ and the magnification is 0.09.

Step 1: Calculate the focal length of the mirror

The focal length is half the radius of curvature.

$$\begin{gathered} \mathrm{f}=‐\frac{\mathrm{R}}{2} \\ \mathrm{f}=‐\frac{6/2}{2} \\ \mathrm{f}=‐1.5\mathrm{cm} \end{gathered}$$

Step 2: Calculate the object distance

$$\begin{gathered} {\mathrm{d}}_{\mathrm{o}}=18‐\mathrm{R} \\ {\mathrm{d}}_{\mathrm{o}}=18‐3 \\ {\mathrm{d}}_{\mathrm{o}}=15\mathrm{cm} \end{gathered}$$ 

Step 3: Calculate the magnification by finding the image distance

The relation between the object distance and the image distance is $\frac{1}{{\mathrm{d}}_{\mathrm{o}}}+\frac{1}{{\mathrm{d}}_{\mathrm{i}}}=\frac{1}{\mathrm{f}}$ .

$$\begin{gathered} \frac{1}{{\mathrm{d}}_{\mathrm{o}}}+\frac{1}{{\mathrm{d}}_{\mathrm{i}}}=\frac{1}{\mathrm{f}} \\ \frac{1}{{\mathrm{d}}_{\mathrm{i}}}=\frac{1}{‐1.5}‐\frac{1}{15} \\ {\mathrm{d}}_{\mathrm{i}}=‐1.36\mathrm{cm} \end{gathered}$$

The magnification is

$$\begin{gathered} \mathrm{m}=‐\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{o}}} \\ \mathrm{m}=0.09 \end{gathered}$$

So, the position is $1.36\mathrm{cm}$ and the magnification is 0.09.