BIO (a) Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed?

Video transcript

Hi, everyone in this practice problem, we're being asked to calculate a teacher's near point. We will have our science teacher wearing a 3.25 diopter contact lens. The options given for the near points are a 31 centimeters from the I B 64 centimeter from the IC, 100 and 33 centimeter from the eye. And lastly D 325 centimeter from the I. So the compact lens function is to form a virtual image at the eyes near point. When an object is placed 25 centimeters away from the eye, the power of the contact lens can be expressed by the uh formula of B equals to one divided by F where F is the focal length. So the focal length when it is being asked to calculate the teacher's near point can then be calculated by uh rearranging the formula where F will then be equals to one divided by 3. or one divided by P. So F equals to one divided by P and then F will equals to one divided by 3.25 diopters. And that will give us an F value or the focal length of 0.3077 m or in centimeter, then F equals to 30.77 centimeter because we want to multiply that by 100 awesome. So now this is not the teachers near point. But in order for us to calculate that we can use the object image relationship formula where one divided by S one plus one divided by S one prime will equals to one divided by F or the focal length. S one is the object distance, one prime is the image distance and F is the focal length. So we want to use this object image relationship for a lens to calculate the image of an object placed 25 centimeter away from I. So that is essentially calculating for S one prime S one prime will determine the teachers near point. So S one prime by rearrangement can then be calculated by multiplying S one with F and dividing that with S one minus or subtracted with F OK. Substituting of this value we have our S one to then be equals to centimeter F to be equals to 30.77 centimeter. S one is given to be 25 centimeter minus that with F which is 30.77 centimeter. This will then give us our S one prime to be negative 133 centimeter. Or essentially the teachers near point is going to be 100 and centimeter from D I just like. So, so 100 and 33 centimeters from the I is going to be the answer to this practice problem which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics available on our website and that'll be it for this one. Thank you.