Question

In astronomy it is common to expose a photographic plate to a particular portion of the night sky for quite some time in order to gather plenty of light. Before leaving a plate exposed to the night sky, Matt decides to test his technique by exposing two photographic plates in his lab to light coming through several pinholes. The source of light is \(1.8 \mathrm{m}\) from one photographic plate and the exposure time is \(1.0 \mathrm{h}\). For how long should Matt expose a second plate located \(4.7 \mathrm{m}\) from the source if the second plate is to have equal exposure (that is, the same energy collected)?

Short answer

Answer: The second plate should be exposed for approximately 6.81 hours.

Step-by-step solution

Understanding the inverse square law

The inverse-square law states that the intensity of light is inversely proportional to the square of the distance from the source. In this problem, we can write this relationship as follows: $$ \frac{I_1}{I_2} = \frac{d_2^2}{d_1^2} $$ where \(I_1\) and \(I_2\) are the intensities of light at the first and second photographic plates, respectively, and \(d_1\) and \(d_2\) are their distances from the light source.

Write the relationship between exposures

To find the exposure time for the second plate, we need to find the relationship between the two exposures. Since the exposure is equal to the intensity of light multiplied by the time, we can write this relationship as: $$ E_1 = E_2 $$ or $$ I_1 t_1 = I_2 t_2 $$ where \(E_1\) and \(E_2\) are the exposures, and \(t_1\) and \(t_2\) are the exposure times for the first and second plates, respectively.

Substitute the inverse square law into the exposures relationship

Now, we can substitute the inverse-square law relationship from Step 1 into the equation from Step 2 to find the relationship between the exposure times: $$ \frac{d_2^2}{d_1^2} I_1 t_1 = I_2 t_2 $$ Simplify the equation by dividing both sides by \(I_1\), to have $$ t_2 = \frac{d_2^2}{d_1^2} t_1 $$

Input the given values and solve for the exposure time

Now we can input the given values: \(d_1=1.8\,\text{m}\), \(d_2=4.7\,\text{m}\), and \(t_1=1.0\,\text{h}\), and solve for \(t_2\): $$ t_2 = \frac{(4.7\,\text{m})^2}{(1.8\,\text{m})^2} \cdot 1.0\,\text{h} $$ After calculating, we get: $$ t_2 \approx 6.81\,\text{h} $$ The second plate should be exposed for approximately \(6.81\,\text{hours}\) to have equal exposure as the first plate.