Exercise a system consists of two lenses mounted on an...

EXERCISE A system consists of two lenses mounted on an optical bench. Lens 1 has focal length 5.80 cm. An object is placed 7.5 cm to the left of this lens. A second lens, Lens 2, with focal length 11.0 cm, is placed to the right of the first lens. A screen is to the right of the second lens. The image on the screen has a magnification of +1.00. What is the distance (in cm) between the object and the screen? Hint First, use the lens equation to find $q_1$, and use this result to calculate $M_1$. Note that $M = M_1M_2 = 1.00$. Use the equation for $M_2$ to find $p_2$ in terms of $f_1$ and $q_2$. Next, use the lens equation to find $q_2$ in terms of $f_2$ and $p_2$. Substitute your expression for $p_2$ into your equation for $q_2$ and simplify. You should get an expression for $q_2$ in terms of $f_1$ and $f_2$, both known quantities. Then, find $p_2$. Finally, find the total distance by summing $p_1 = 7.5$ cm, $p_2$, $q_1$, and $q_2$. Click the hint button again to remove this hint. 29.13 x cm

Transcript

00:01 In order to solve the first half of this question that states, bonus one question, you are analyzing the ages of students in the sociology class.

00:10 The ages and years of 10 randomly selected students are as follows 19, 21, 20, 22, 18, 23, 19, 20, 22, and 21.

00:20 And you're told to calculate the standard deviation of all the ages in this sample of students.

00:24 We are going to need to calculate the standard deviation of the data that we were given here.

00:32 So we'll be using this formula and the first thing that we're going to need to do is calculate average value of these 10 digits here, otherwise known as the arithmetic mean.

00:47 In order to find the average value, we are going to need to add up all the values 19 plus 21 plus 20 plus 22 plus 18 plus 23 plus 19 plus 20 plus 22 plus 21 equaling 205 and then divide by how many numbers that we had added up.

01:05 That's to say 10 numbers were added up, so 205 divided by 10 equals 20 .5, sorry, 20 .5 as an average.

01:14 X refers to each one of these individual numbers on a come -by -come basis, and x average refers to this 20 .5 value that we just calculated.

01:23 Now we must calculate the difference between these two values, that is to say, each value minus the average.

01:31 For example, 19, which will be x in this case, minus 20 .5, which should be x average, equaling negative 1 .5.

01:39 And we will do this down the list of numbers until we have all of them.

01:42 Then we are going to take all those values, and we're going to take the root, the square them.

01:49 We are going to square them, and that will look like negative 1 .5 squared equaling 1 .5, negative 1 .5 times negative 1 .5, because squaring is just multiplying the number by itself, and it has the property of negative numbers multiplied by themselves, turn into positive numbers, so every single one of our values is going to be a positive number.

02:11 And these values are what are going to be inside this parentheses here.

02:15 And this symbol here means the sum.

02:17 That is the next step in this process.

02:20 We're going to be taking all of these squared values and adding them together.

02:24 0 .25 plus 0 .25 plus 2 .25 plus 6 .25 plus 6 .25 plus 6 .25 plus 2 .25 plus 2 .25, plus 2 .25, plus 2 .5, plus 2 .5 .5, plus 2 .5 .5, plus 2 .25 plus 0 .25 plus 0 .25 equaling 20 .25, which will replace this entire numerator section of this section here.

02:47 And then we will be left with 20 .25 divided by n minus 1, where n is equal again to the number of terms that we had 10, where n minus 1 would be 10 minus 1 or 9.

03:01 So we get 20 .5, 20 .25...

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