A farsighted person has a near point of \$50 \mathrmcm\$. What strength lens, in diopters, is necessary to bring his near point to \$25 \mathrmcm ?\$

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Video Transcript

Okay, for this reason in this problem, we have a much sighted person with a near suggest off 50 centimeters. Then we need to use a lens to bring this near point 2 25 ST T meters. Okay, on. We desire to recognize what need to be the strength off this strength, the toughness of this lands. For this reason the toughness of the lens or the power off the lens doesn't matter the word. It's usually one separated by the focal lunch. And the devices for this is the di Achter. Okay, so we desire to understand what have to be the focal floor or the power, the toughness of the floor to change the near allude of a human being over first sighted person. However let's recognize that the near suggest is the closest suggest where this person deserve to have have the right to see the object v clarity. Therefore is the focus. Is the limit off the focal allude of this person? Ah, and also a normal near suggest for our human is around 25 centimeters. Okay, therefore to calculation one separated by F, the focal the focal distance length. We currently know the equation the we need to use because it's the standard equation for lens is the soil equation. So one split by F equates to one separated by zero to add one, divided by T I. Therefore is the distance off the object and also the distance off the image? Okay, yet what must be the street of the object and what should be the street off the picture in this particular case? for this reason we desire to kind the image, the thing in the 25 Let's put here, we want the object and 25 centimeters. Therefore we want to carry the object for 25. However there is a problem. We must pay attention because when we have actually ah, glass in front of ours eyes, these distance right here needs to it is in subtract by the distance between the eyes and also the lens. Okay, therefore, the distance off the object in this particular case is 23 0.5 1.5 is the traditional distance between the lens, the glasses and also the i would of a person. Okay, we have the right to assume these distance as 1.5 and doing the very same for the street off the image, which is in 50. Us subtract 1.5 and we gain 48.5. Currently we're going to calculate utilizing this numbers, but this was extremely important to fix this problem. We require to consider the distance in between the eyes and also the lens since the lens equation is not about the distance between the I and also the thing is around the distance between the lens and the object. That's why we must subtract this value from 25. Okay, so coming back to the equation, Well, this is one divided by F amounts to 10 square, divided by 48.5 add to 10 square, divided by 23.5. Okay, calculating this, we discovered that, or strength or strength, i beg your pardon is one separated by F it's walk to be two suggest 19. The doctors.

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Okay, therefore that's the final answer come this problem. Many thanks for watching.