F1 60 d1 100 d2 40 f3 75 d3 40 10 20 for an object at 400 mm...

$$f_1 = 60$$ $$D_1 = 100$$ $$D_2 = 40$$ $$f_3 = -75$$ $$D_3 = 40$$ 10 20 For an object at 400 mm to the left of lens 1. Make an accurate sketch of the system including at least the marginal and chief rays and several other rays that seem relevant to you. What is the diameter of the field of view in image space? Express your answer in mm with three significant figures. Enter answer here What is the angular field of view in image space? Express your answer in radians with three significant figures. Remember that this is the angle from the axis to the chief ray, so corresponds to half the total angular range in the field. Enter answer here

Transcript

00:01 For our question, we have three lenses with focal lengths f1, f2, and f3.

00:05 The distance from the object to the first lens is p1, and the distance is between lens 1 and 2 is d sub 1, 2.

00:12 And the distance between lenses 2 and 3 is d sub 2.

00:15 For part a, we're asked to find the distance to the image produced by the third lens.

00:21 So first we have to find the distance to the image produced by the first lens.

00:25 That's the object of the second lens.

00:27 Then we have to find the distance produced by the third lens.

00:30 The distance of the image produced by the second lens, because that's the object of the third lens, then we can find the distance produced by the third lens, or the image produced by that third lens.

00:40 So first we're going to find i sub 1.

00:41 To do this, we're going to use the small lens equation, which says 1 over i sub 1, plus 1 over p sub 1 is equal to 1 over the focal length, f sub 1.

00:53 I sub 1 is the distance to the image of the first lens, so we're going to solve for that.

00:57 We find that i sub 1 is equal to 1 over f sub 1 minus 1 over p sub 1, all raised to the negative 1.

01:09 We have to raise it to the negative 1 to get i sub 1 out of the denominator.

01:14 We find that this is equal to 9 centimeters.

01:18 So the distance from the object of the second lens to the second lens, which we call p sub 2, is the distance between lens 1 and 2, d sub 1, minus i sub 1.

01:30 Plugging in that value we find that this is equal to 6 centimeters.

01:38 Now we can calculate i sub 2 the same way we calculated i sub 1, but we simply replace f sub 1 with f sub 2 and p sub 1 with p sub 2.

01:54 Plugging in those values we find that i sub 2 is equal to 6 centimeters.

02:02 Continuing, we now can calculate p sub 3, the distance from the object of the third lens to the third lens, and this is d sub 2, 3, the distance between lens 2 and 3, minus i sub 2.

02:16 And this comes out to equal 5 centimeters...

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