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A person whose near-point distance is 42.5 $\mathrm{cm}$ wears a pair ofglasses 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 thefocal length and refractive power of her glasses.

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

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Rutgers, The State University of New Jersey

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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'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're going to make the same the distance off the image. It is the near point or the fair point. Let's see in your point, she'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'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's the answer to this problem. Thanks for

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