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As usual, show all algebraic steps

Image Link:http://www.colorado.edu/physics/phys2020/phys2020_fa15/Written%20HW/Physics%202020%20HW%2010.pdf

******* I only need help with Question 5! I put questions 3 and 4 up so you could see my answer for them because I think you might need that information to complete question 5*****

Question 3. Suppose we want to use a set of eyeglasses to allow this eye to read comfortably at a distance of 25 cm. Suppose the lens of the eyeglass is 1 cm from the front of the lens of the eye. In this situation, the object (the book) is now at 25 cm from the eye. The eye glasses should produce an image at 40 cm from the eye. So we are now considering the lens of the eyeglasses

Part A) For the eyueglass lens, what is the object distance?

ANSWER: 24 cm

Part B) Notice the image will be one the same side as the object. What is the image distance relative to the eyeglass lens, and is it negative or positive?

ANSWER: - 39 cm

Question 4. Using the image distance and object distance from above, calculate the focal length of the eyeglass lens. (You should get +62.4 cm). Is this lens a converging lens or a diverging lens?

ANSWER: + 62.4 cm (converging lens since f is positive)

Question 5: .The image produced by the eyeglasses now becomes the OBJECT for the lens of the eye. Using your result for the focal length of the eyeglasses, assume the person wearing glasses looks at an object 30 cm from the eye. What will the focal length of the eye need to be now to get a sharp image on the retina? Answer in mm to 1 decimal place. Using the information given, you will first want to calculate the image distance from the eyeglass lens, and then use that distance to get the object distance for the lens of the eye. Remember, the lens of the eye is 1 cm from the eyeglasses, and an object distance on the left side of these lenses will be positive. Draw a diagram.

Explanation / Answer

5.)
Given - The image produced by the eyeglasses now becomes the OBJECT for the lens of the eye.
Therefore -
For the Lens of eye -
do = 40 cm = 400 mm
The image needs to be formed at Retina -
di = 17 mm

1/f = 1/di + 1/do
1/f = 1/400 + 1/17
f = 16.30 mm

focal length of the eye need to be now to get a sharp image on the retina , f = 16.30 mm

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