Questions: The exposure index EI for a camera is a measurement of the amount of light that hits the image receptor. It is determined by the equation EI=log2(f^2/t), where f is the "f-stop" setting on the camera, and t is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be?

The exposure index EI for a camera is a measurement of the amount of light that hits the image receptor. It is determined by the equation EI=log2(f^2/t), where f is the "f-stop" setting on the camera, and t is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be?

Transcript text: The exposure index $E I$ for a camera is a measurement of the amount of light that hits the image receptor. It is determined by the equation $E I=\log _{2}\left(\frac{f^{2}}{t}\right)$, where $f$ is the "f-stop" setting on the camera, and $t$ is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be?

Solution

Solution Steps

To find the exposure index $ EI $, we need to substitute the given values of the f-stop setting $ f = 8 $ and the exposure time $ t = 2 $ seconds into the formula $ EI = \log_2\left(\frac{f^2}{t}\right) $. We will then calculate the logarithm base 2 of the resulting fraction.

Step 1: Substitute Values

We start with the formula for the exposure index: \[ EI = \log_2\left(\frac{f^2}{t}\right) \] Given $ f = 8 $ and $ t = 2 $, we substitute these values into the equation: \[ EI = \log_2\left(\frac{8^2}{2}\right) \]

Step 2: Calculate the Fraction

Next, we calculate $ 8^2 $ and then divide by $ 2 $: \[ 8^2 = 64 \quad \text{and thus} \quad \frac{64}{2} = 32 \]

Step 3: Calculate the Logarithm

Now we compute the logarithm base 2 of $ 32 $: \[ EI = \log_2(32) \] Since $ 32 = 2^5 $, we find: \[ EI = 5 \]