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The contact lens prescription for a mildly farsighted person is 0.750 D, and the person has a near point of 29.0 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?
$P_{\text {tear }}=-0.198 \mathrm{D}$
Physics 103
Chapter 26
Vision and Optical Instruments
Reflection and Refraction of Light
Wave Optics
University of Washington
University of Sheffield
University of Winnipeg
McMaster University
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while afford a normal near object can be found by Pete one that is a cool thio 100 derided by the focal point. So we'll find a focal. And here, so the focal and can be found for a year. Object is one where f is a quote to one or extends to the image plus one or distance to the object. Oh, then substituting the values we get here one of wherever. He's a cool Jew. One over 2.5, 2.5 centimetre for a distance. The image plus one award 25 to the distance of objects and then we get eh, Fred, you hear? He's a 1.1 point 85 85 So substituting this value off F or our P one we get here 100 divided by 1.858 Trife. The issues are so the power off for the normal near object is 54 doctors Now for the person for the far we have. Um um, one word, if that is equal to one word D I, which is 29 centimeters plus one were to food deals. Object. This gives us the focal end. Um f is a cool too. 1.87 centimeter. Substituting this in the formal. Oh, for the power we get 100 divided by 1.87 Uses the power off 50 53.4 doctors. So the power off the tier will be powered off. The tier will be 54 minus but one personally 53.4 plus, uh, 0.75 given. Oh, this use us A minus minus 0.154 doctors. So this is the power off the tier.
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