a-person-whose-near-point-distance-is-425-mathrmcm-wears-a-pair-of-glasses-that-are-21-mathrmcm-from

Question

A person whose near-point distance is 42.5 $\mathrm{cm}$ wears a pair of
glasses that are 2.1 $\mathrm{cm}$ from her eyes. With the aid of these glasses,
she can now focus on objects 25 $\mathrm{cm}$ away from her eyes. Find the
focal length and refractive power of her glasses.

Nathan Silvano · Accepted Answer

Okay, So in this problem, we have a person with a near point distance off 42.5 centimeters. And this person, whereas a pair of glasses are two point once intimidators from her eyes. Okay, we want to discover what needs to be the focal lens. So this person I can see clearly objects. There are 25 centimeters from his eyes. So to calculate Focal Lent, we always use the feelings equation, which is one divided by zero. Well, that&#x27;s one divided by the I. All this to the power off minus one. And first we need to discover what is the zero distance off the object from D bless. So the zero is always the distance of the object to the glass. We know that the distance off the object to the eye is 25. If we subtract the distance off the less d I just 2.1, we have a distance off the off the object to the glass and the I. We&#x27;re going to make the same the distance off the image. It is the near point or the fair point. Let&#x27;s see in your point, she&#x27;s 42 0.5. My nurse 2.1 in the distance off the glass to D IE. So if we put these values in the formula, we&#x27;re just going to have one divided by 25 minus 2.1 plus 42.5 minus 2.1 with a minor signing front. Because this is a near point distance. All this the power of minus one. So finally, the focal lens off these glasses needs to be 53. Saying teenagers and the refractive power we know how to calculate also their refractive power fortune. Here our is just one divided by 50 tree riches, one point nine the adopters. And that&#x27;s the answer to this problem. Thanks for

Dading Chen · Answer

All right. So the question was asking us what should be the power of the lances. This is four glasses. Well, we know that the image distance equals inactive. 17 point does. And since that, you were saying that the students Farpoint is 17 centimeters. So that&#x27;s why the value it&#x27;s not too. Okay, so, um, the option distance is equal to 45.0. Centimeters is the distance between a computer screen to her glasses, which is point for five years. So we know that the focal s can be equal to your V over you possibly be. And if we plug in a value, we&#x27;ll get it you, which is just 0.45 meter times. Negative point wants Have a meter over 45. I mean, sorry. Point full. If I meter plus negative point, uh, 17 meter and this will give us the focal s a swell. Connected on 27 meter. So now we know the focal lanes open this question so we can figure out the power. What is question? Which is he? He was won over half. Okay. Which can be equal to one over Joe. Well over 0.27 meter. I mean, find this 0.2, Samir, and this will give us next it three points. 70 Peter to the power of negative one, which is naked. 3.70 The overtures. Okay. And you can see, since the power is connected, that means that the lenses she wears is diversions. Dances. Okay, so that&#x27;s the solution. Parties question. Thank you.

Adnan Gill · Answer

So the power of cleanse is giving us negative 1.5 years. And what that means is that one over F is equal to negative 1.5 issue, and the near point for the person is your 0.0.69 meters, which then means that the I will be equal to negative off 0.69 year old minus D, where D is the distance between her I and violence. And then, since we&#x27;re talking about focusing distant optics, a d o would be equal to infinity, which will tend me that one or g o, is equal to zero. Now we can use the Finland&#x27;s Formula One over Dio plus one over the I is equal to one over f So one over dio is you. Also, we can eliminate that and then one or negative. As you don&#x27;t 0.690 my SD is equal to negative 1.5. So the negative and the negative when both sides are negatively can report, say this positive and work. But then that give us is 1.5 times is your 0.69 year old minus D is equal to one or a 0.690 minus D is equal to 0.667 and then D will tell people to 0.69 year old minus 0.667 which comes out in different was zero point 0 to 3 meters.

Netra Sharma · Answer

so a near sighted person has a near point at 16 centimeter and Farpoint at 25 cent. Vera. So in part A. If we want to find the correcting glance that are placed two centimeter in front of the persons, I then for a near sighted person to see the distant objects. That means when P is he called infinity, the image should form at fire Point, and it is virtually an upright. So the Farpoint image distance should be negative. 25 centimeter, right? But here&#x27;s the thing, since the lenses place two centimeter in front of the eyes is actually going to be. The aim is distance is actually going to be 20 negative, 25 minus to subtract that and this is going to be negative. 23 centimeter. Okay, so the focal lens in this case when the object distances infantry is basically F because Q because negative 23 centimeter. All right, I&#x27;m just gonna write down this one in terms of meter. Is there a point to three meter? So from here, the power of the lens, which is Q. They&#x27;re going to one of her focal length in matter, and this is going to be won over negative 0.23 This will give me negative 4.3 doctors. So we do not need correcting lens for the near point because the person is nearsighted and the normal near point of ah, regular normal near point off a person usual normal nor near point is 25 centimeter. But the person has the near point less than 25 centimeter. So we don&#x27;t really need a correcting less for the your point for that person. Okay, so let&#x27;s go to party now. If the contact lenses are used instead of my glasses, then your part B when the object is at infinity, when P equals infinity Q instead off negative. 23 centimeter is gonna be now, actually, negative 25 centimeter and the negative sign. I&#x27;m gonna mention that again. The negative sign is because the virtual a practice Mrs formed at No, the fire point of the person. Okay, so from here, the focal length again f is going to be called to Q, which is negative 25 sending meter and this is negative 0.25 meter. So the power of the contact list the person has to use has to be one of her f in meters, which is one over negative. 0.25 And this will give me negative four doctors. Okay, negative. Four adapters for this case.

Nathan Silvano · Answer

Okay, so and this probably have ah, first sighted person wearing glasses with a reflective power put in here. Effective power off tree point six die up hers. The glass is created an image of the person near point off on the object that is 25. Same team enters. We can call this the zero, and we want to calculate what is the near point of this person and near point when this person was not using glasses. Okay, so first of all, let&#x27;s calculate the distance off image where the images produce by these glasses. So we have the Finland&#x27;s equations for the distance of the image that is just one divided by af miners, one divided by zero onto the power of minus one. And we know that one divided by F is the same as the refractive power. So this is just reflective. Our miners one divided I d zero all to the power of miners one. So this distance distance of the image produce is going to be 3.6 miners, zero point 25 which is the distance is, you know, minus 0.0 25. What is this number? This number is the distance off the glass. So the eyes of the person, because the near this distance off 25 cent emitters is related to the eyes of the person and we need to relate this distance to the glasses. So we have to subtract. Bye. 0.0 25 All this student power of minus one. So finally we have this distance off 1.18 meters. Okay, now we can calculate in near the near point distance of this person without the glasses, because the near point distance is just one point 18 plus the distance zero point 25 which gives us one point 21 meters. And that&#x27;s the answer to this problem. Thanks for watching.

Nathan Silvano · Answer

Okay, So in this problem, a person needs reflective power off. Let&#x27;s see. Were effective. Our one off minus 0.44 five two farther distance and our effective power. That&#x27;s car effective. Powered too off one point 85 Diop tres I forgot to put in here. This is a five day adopters. Also 1 85 Diop tres was a thief. Okay, so we know that this is a reflective power to a diverging type of lens, and this one is for a home verging type of lens. Let&#x27;s begin with the 1st 1 when in this problem, we need to calculate it the distance off the near and far points of this person uncorrected vision. And we also know that the near point of this person is 25 centimeters, which is the standard, and the glasses at a distance of two centimeters from the ice. Okay, so let&#x27;s begin with the 1st 1 We know that the first reflective power is for a diverging type of lens, and we know that the diverging type of lens it&#x27;s used to correct the near sighted distance, the near sight problem. And we know that in the near side problem. The person cannot see objects there. There they are infinity. So too calculated the distance of the image that we want. We just need to use the finger lands a creation one divided by half miners, one divided by zero Oh, to the power of minus one. We know that the refractive power, by definition is just one divided by F. So here we have miners zero point for 45 miners, one divided by infinitive and one divided by infinity is zero. So we just have in here miners to a point 25 meters. And that&#x27;s the distance off the image produce by the glasses of this person. But remember that we do not want the distance worth the image produced by the glasses. We want the distance that a person can see without the glasses. So the Farpoint, this person, you&#x27;re far weight of this person just going to be the absolute value off distance. You can see that it&#x27;s 2.25 meters, some of with distance off the glasses, which is, you know, point did too. So this person can actually actually see without the glasses a distance off two point 27 meters. Beyond that, this person needs a glass to correct his vision. Okay, now that&#x27;s calculator did near point again. We know that the distance off the image for by this glasses it&#x27;s going to be one divided by F miners, one divided by he&#x27;s zero oh, to the power of minus one. The refractive power off the converging type of lens is one point 85 minus the distance off the object. The distance of the object needs to be the distance of in your point, which is 0.25 miners. The distance off the glasses, which is Europe, wants you to to the power of minus one. That&#x27;s equal to minus zero point for meters. Okay, so the near point near point of this person without the glasses again is the absolute value he can see if the glasses lost the distance off the glasses richer, which is, um, 0.2, actually general 0.4 meters. So let&#x27;s put in order unities. It is going to be 40. Same team enters, we lost two cent emitters equals two 42. Same team enters. So this is the near point off this person without the glasses. That&#x27;s the final answer to the problem

Problem

A pair of eyeglasses is designed to allow a perso…

Problem 36 Medium Difficulty

A person whose near-point distance is 42.5 $\mathrm{cm}$ wears a pair of
glasses that are 2.1 $\mathrm{cm}$ from her eyes. With the aid of these glasses,
she can now focus on objects 25 $\mathrm{cm}$ away from her eyes. Find the
focal length and refractive power of her glasses.

Answer

52.9 $\mathrm{cm}$
$+1.89 \mathrm{D}$

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Video Transcript

James S. Walker

Physics

Christina K.

Zachary M.

Aspen F.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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52.9 $\mathrm{cm}$$+1.89 \mathrm{D}$

Physics

Christina K.

Zachary M.

Aspen F.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Christina K.

Zachary M.

Aspen F.

Jared E.

52.9 $\mathrm{cm}$
$+1.89 \mathrm{D}$

James S. Walker

Physics