Q7E - University Physics with Modern Physics

Question

The diameter of Mars is 6794km, and its minimum distance from the earth is $\mathbf{5}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{k}\mathit{m}$. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75m.

Step-by-Step Solution

Answer

The diameter of the image is 0.213mm .

Step 1: Calculate the distance of the image from the mirror

Given that the diameter of the mars is $y=6794km$and the minimum distance is $s=5.58\times {10}^{10}m$.

The relationship between the object distance and the image distance is

$\frac{\mathbf{1}}{\mathbf{s}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

So,

$$\begin{gathered} s\text{'}=\frac{sf}{s-f} \\ s\text{'}=\frac{5.58\times {10}^{10}\times 1.75}{5.58\times {10}^{10}\times 1.75} \\ s\text{'}=1.75m \end{gathered}$$

Step 2: Calculate the diameter of the image using lateral magnification

The lateral magnification is given by $\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathbf{s}\mathbf{\text{'}}}{\mathbf{s}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}\mathbf{\text{'}}}{\mathbf{y}}$

$$\begin{gathered} m=-\frac{1.75}{5.58\times {10}^{10}} \\ m=-3.14\times {10}^{-11} \end{gathered}$$

Now, use the lateral magnification to calculate the diameter of the image

$$\begin{gathered} \left|y\text{'}\right|=\left|my\right| \\ \left|y\text{'}\right|=\left|-3.14\times {10}^{-11}\times 6794\times {10}^3\right| \\ \left|y\text{'}\right|=0.213mm \end{gathered}$$

Hence, the diameter of the image is $0.213mm$.