the-focal-length-for-light-that-strikes-near-the-center-of-a-spherical-convex-lens-is-15-mathrmcm-re

Question

The focal length for light that strikes near the center of a
spherical convex lens is 15 $\mathrm{cm} .$ Referring to Figure $27-23,$ will
the focal length for light that strikes near the edge of the lens be
greater than, less than, or equal to 15 $\mathrm{cm} ?$

Nathan Silvano · Accepted Answer

Okay, so in this questions we have that the light rays they strikes near the center office. Vertical convex lens is 15. Think team enters, so we have the four co land off 15. Same team enters for the light race. There he is, near the center. So if we make a schematic drawing drawing here, we can see that the light races this strike near They grow virg toe a focus off 15. This is the position. 15 same team mentors, but the light race. They strike the lens near the edge, right on the edge of the lens. They have a much near Jeff, so they&#x27;ve the focus off this light race, but much closer. So the focus. It&#x27;s a smaller than the focus off the light rays, the hit near the center of the convex lens. So the answer to this problem is that let&#x27;s go F B, which is the focal lens off the light raise. The heat on the edge of the Comex lands is less then 15. Okay, always remember the half off the light race. And then, when the heat the edge e r deflect much more than the light rays hit in the center of the off the lens

Nathan Silvano · Answer

Okay, so I make a draw off the figure that we have in this problem. And we have our diverging lens and a light rate that divide and a blue wavelength in a rad Waveland. And we can see that these to Waveland has different focal points because they diverge with different angle angles. So let&#x27;s call the focal point of blue and the focal point of red. Okay, so what? We want to discover the problem. The problem say&#x27;s that the focal focal length off the red is 15. Same team enters, and we need to say, if the focal lens off the wave off the blue Wave land is going to be less or greater than 15. So as we can see because Ray that verges away from the focal point, it is clear that the focal point off the blue light is closer to the lens, then the focal point for the the red light. Because if we so, if the focal point of the loot the blue is closer, then we can say that the answer to this problem, it&#x27;s going to me. The focal point of Blue is less then 15 senti mentors, and that&#x27;s the answer to this problem. Thanks for watching

Mayukh Banik · Answer

focused and is given by one. We&#x27;re F universe. It was and do minus and one minus one in worse when were are one minus one over R do in inverse. So now we get put the numbers in and do over anyone is one point 5/1 minus one in the worse are one is infinity minus one over are bleak was negative 35 centimeter with the indoor sign So from here we find that f equals 70 centimeters.

Mayukh Banik · Answer

here. We&#x27;re trying to find the AMS Distin&#x27;s D I. It was one of our F minus one deal. In words of that, it was one of over 20 centimeters, minus 1/25 centimeter inwards off that Equus 100 sent.

Mayukh Banik · Answer

First of all in this problem, you should identify that object distance. It was twice the focal Lent optic distance. He&#x27;s 34 centimeter and focus then is 17 centimeter. Just by looking at it, you should know that Amos Justin should also be 34 centimeter. But this well be clear from the raid item as well. So here we have a chance. Ah, because axis, that is f and U F Over here, we have got f and two f. Here is where we have the candle that&#x27;s used blue for the red eye drum. We were just to raise one para the to the principal, the access and the other ray going through the focus The light that is going through the focus would be proud of to the pick of the axis. The light that was parable Initially, the will go through the focus and we&#x27;ll go in this direction. Uh, they did not come out that well. The line has to be straight like that. So now you would have your object. Exactly. At bless. So this is your object Distance. 34 centimeter is the Oh, this is your emails distance which came out also to be 34 centimeter. Ask previously

Ashley Jordon · Answer

with the information they gave us. We know her focus is at three comma zero someone&#x27;s RPI value is three. So when I plug that in to this equation four times three is 12. So that leaves us with X square equals 12 live. When I grabbed this, this is what our graph ends up looking like. Notice how I have my focus. I also have the plaint zero comma 00.5 plotted because we want to know how wide are graph is when our focal point is that half a centimeter away. Remember text. So I&#x27;m trying to find the distance here. So what we can do to find those X values is plug in 0.5 for why and to our equation. So X squared equals 12 times 120.5 x square equal six Undo a square. Take the square root of both sides were lacked with X equals plus or minus route six. So that means that is the expat or the Y value. Here is that 0.5 The X value is at negative six and for this point were at route six comma point by, So the square root of six is about 2.45 So if I add 2.45 plus 2.45 together we get 4.9. We&#x27;re talking about centimeters here, so that is the width of this lens when our focal point is half a centimeter away.

Problem

The focal length for red light that strikes a sph…

Problem 79 Medium Difficulty

The focal length for light that strikes near the center of a
spherical convex lens is 15 $\mathrm{cm} .$ Referring to Figure $27-23,$ will
the focal length for light that strikes near the edge of the lens be
greater than, less than, or equal to 15 $\mathrm{cm} ?$

Answer

less than 15 $\mathrm{cm} .$

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less than 15 $\mathrm{cm} .$

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Christina K.

Andy C.

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James S. Walker

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