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University of Wisconsin - Milwaukee

# The accommodation limits for Nearsighted Nick’s eyes are 18.0 cm and 80.0 cm. When he wears his glasses, he is able to see faraway objects clearly. At what minimum distance is he able to see objects clearly?

## 23.2 \mathrm{cm}

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

### Video Transcript

in this problem. A person is near sighted and has a near point at 18 centimeter and Farpoint at 80 centimeter. And he works a pair of glasses. And when he wears a pair of glasses, he's able to see fire away objects Very clearly, which means when P, which is the object distance, is at infinity, he's able to form the image. Ah, the Farpoint, which is negative. 80 centimeter. The reason why you see the negative sign here is because the image is a pride and virtual and actually in front of the lens. So from this information, we can actually find the focal length of the less he's using to find F. We just use less Formula One of her F because whatever p plus one over a cue and one of her p is one of her infinity, one over negative 80 one of our infinity is zero. So if you just take the risk, it'll trickle of whatever f it is gonna be F because negative a decent debater. Sorry, that's gonna be negative. 80 Cindy Veeder And if he wants to be able to see an object at a close distance, there's actually a minimum limit to that, and in that case, the image distance will be at the near point. All right, so let's formulate that. So in this case, when he wants you to get the closer object, the immense distance would be at negative 18. The same reason the object is the image is gonna be, firstly, bitch up pride and in front of the lens. And for this case, we want to find out the minimum uptick distance. So I will call the immense distance as Q prime and Object Distance P prime. So in this case, P prime. If you want to use the less formula, it is basically going to be Q prime times. Focal late, which we know from the part A of the problem over Q minus F and Q Prime is negative. 18 times negative, 80 over negative 18 minus minutes plus 80. So this will give us 23.2. Said it better

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics

##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg