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Chapter 33: Problem 13

When a diver with \(20 / 20\) vision removes her mask underwater, her vision becomes blurry. Why is this the case? Does the diver become nearsighted (eye lens focuses in front of retina) or farsighted (eye lens focuses behind retina)? As the index of refraction of the medium approaches that of the lens, where does the object get imaged? Typically, the index of refraction for water is 1.33 , while the index of refraction for the lens in a human eye is \(1.40 .\)

Short Answer

Expert verified
Answer: When a diver removes their mask underwater, their vision becomes nearsighted (myopic) because the light converges less when it passes through water, which has a different index of refraction than air. This results in the lens focusing light in front of the retina.

Step by step solution

01

Understand the Principle of Refraction and its Effect on Vision

Refraction is a phenomenon in which light changes direction when it passes from one medium to another, with differing indices of refraction. This change in direction causes the light to bend, which can impact how the lens in the eye focuses the incoming light on the retina. In the case of the diver, the refraction of light is altered due to the change in medium as the light passes through the water and enters the eye.
02

Determine Whether the Diver's Vision Becomes Nearsighted or Farsighted

When wearing the mask, there is a layer of air between the diver's eye and the water. This allows the eye to focus properly on objects underwater. When the mask is removed, the eye is directly in contact with water, which has a different index of refraction than air. The diver's eye lens must now adjust to focus on objects in the water, with its index of refraction 1.33. Comparing the index of refraction of water which is 1.33 to the human eye's lens index of refraction, which is 1.40, the light converges less when it passes through water than through air. This results in the lens focusing light in front of the retina, making the diver myopic, or nearsighted.
03

Discuss Image Formation as the Index of Refraction of the Medium Approaches that of the Lens

As the index of refraction of the surrounding medium (water) approaches that of the lens (1.40), the light bending at the lens's interface decreases. As a result, the power of the lens to focus light decreases, causing the image to be formed at a greater distance from the lens. When the index of refraction of water becomes equal to that of the lens, the lens will lose its ability to focus the light, and the object will be imaged at infinity, making it impossible to focus on nearby objects.

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Most popular questions from this chapter

An object is placed on the left of a converging lens at a distance that is less than the focal length of the lens. The image produced will be a) real and inverted. c) virtual and inverted. b) virtual and erect. d) real and erect.

The focal length of the lens of a camera is \(38.0 \mathrm{~mm}\). How far must the lens be moved to change focus from a person \(30.0 \mathrm{~m}\) away to one that is \(5.00 \mathrm{~m}\) away?

A camera has a lens with a focal length of \(60 . \mathrm{mm} .\) Suppose you replace the normal lens with a zoom lens whose focal length can be varied from \(35 . \mathrm{mm}\) to \(250 . \mathrm{mm}\) and use the camera to photograph an object at infinity. Compared to a 60.-mm lens, what magnification of the image would be achieved using the \(240 .-\mathrm{mm}\) focal length?

Several small drops of paint (less than \(1 \mathrm{~mm}\) in diameter) splatter on a painter's eyeglasses, which are approximately \(2 \mathrm{~cm}\) in front of the painter's eyes. Do the dots appear in what the painter sees? How do the dots affect what the painter sees?

A telescope has been properly focused on the Sun. You want to observe the Sun visually, but to protect your sight you don't want to look through the eyepiece; rather, you want to project an image of the Sun on a screen \(1.5 \mathrm{~m}\) behind (the original position of) the eyepiece, and observe that. If the focal length of the eyepiece is \(8.0 \mathrm{~cm},\) how must you move the eyepiece?
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