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University of Wisconsin - Milwaukee

# A particular nearsighted patient can’t see objects clearly beyond 15.0 cm from their eye. Determine (a) the lens power required to correct the patient’s vision and (b) the type of lens required (converging or diverging). Neglect the distance between the eye and the corrective lens.

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield

### Video Transcript

in this problem, a near sighted person has a fire point of 15 centimetre because he can't see beyond 15 centimetre. Okay, So to find the corrective linds power, remember that the virtual and upright images form at the fire point when the object instances that infinity Okay, so let's start. Let's find the power for the less. So when the object distance is at infinity, the emits distance has to be at negative 15 centimetre. All right, so I'm just gonna write down in terms of meter negative. 5th 15 centimetre is going to be negative 0.15 meter. So from here, if you want to use the lens equation to find the power power P is going to be gold to one of her F, which is one of her P plus one of her cue and one of herpes one over infinity, which is going to be zero. And this is going to be won over minus 0.15 in meters. So this will give you negative 6.66 dieters. Okay, so for pervy for part B here since the power of the lens is negative, the lens that is used for the correction is actually a diverse England's

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics

##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

LB
##### Aspen F.

University of Sheffield