Question

The main mirror of a telescope has a diameter of \(4.713 \mathrm{~m}\). The mirror is made of glass with a linear expansion coefficient of \(3.789 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) By how much would the temperature of the mirror need to be raised to increase the area of the mirror by \(4.750 \cdot 10^{-3} \mathrm{~m}^{2}\) ?

Short answer

Answer: The temperature of the mirror needs to be raised by approximately 89.08°C to increase the area of the mirror by 4.750 x 10^-3 m^2.

Step-by-step solution

Initial area of the mirror

Let's first find the initial area of the mirror using the given diameter. Since the mirror is circular, we can use the formula for the area of a circle: \(A = \pi r^2\), where \(A\) is the area and \(r\) is the radius of the mirror. The radius is half of the diameter, so: \(r = \frac{1}{2} \times 4.713 \mathrm{~m} = 2.3565 \mathrm{~m}\). Now we can find the initial area: \(A_{initial} = \pi (2.3565)^2 = 17.452 \mathrm{~m}^{2}\) (approximately).

Final area of the mirror

As we have the desired increase in the mirror's area, we can calculate the final area by adding this increase to the initial area: \(A_{final} = A_{initial} + \Delta A = 17.452 + 4.750 \cdot 10^{-3} = 17.4567\mathrm{~m}^{2}\).

Linear increase in radius

We need to express the change in the area of the mirror in terms of the linear increase in radius. To do so, let's denote the final radius of the mirror as \(r_f\). The final area of the mirror can be expressed in terms of the final radius as follows: \(A_{final} = \pi {r_f}^2\). Now, we can find the final radius: \(r_f = \sqrt{\frac{A_{final}}{\pi}} = \sqrt{\frac{17.4567}{\pi}} = 2.35713\mathrm{~m}\) (approximately). The linear increase in radius is: \(\Delta r = r_f - r = 2.35713 - 2.3565 = 6.3 \times 10^{-4} \mathrm{~m}\) (approximately).

Change in temperature

Now we can use the linear expansion coefficient and the change in radius to find the temperature change needed for the expansion. Recall the formula for linear expansion: \(\Delta L = L \alpha \Delta T\), where \(\Delta L\) is the change in length, \(L\) is the initial length, \(\alpha\) is the linear expansion coefficient, and \(\Delta T\) is the change in temperature. In this case, \(\Delta L\) corresponds to \(\Delta r\), and \(L\) corresponds to the initial radius of the mirror. Therefore, we have: \(\Delta r = r \alpha \Delta T\). Now, we can solve for the change in temperature: \(\Delta T = \frac{\Delta r}{r \alpha} = \frac{6.3 \times 10^{-4}}{2.3565 \times 3.789 \times 10^{-6}} = 89.08^{\circ}\mathrm{C}\) (approximately). So, the temperature of the mirror needs to be raised by approximately \(89.08^{\circ}\mathrm{C}\) to increase the area of the mirror by \(4.750 \cdot 10^{-3}\mathrm{~m}^{2}\).