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Chapter 24: Problem 62

(a) If Harry has a near point of \(1.5 \mathrm{m},\) what focal length contact lenses does he require? (b) What is the power of these lenses in diopters?

Short Answer

Expert verified
In this exercise, we are given the near point distance of Harry's eyes (1.5 meters) and asked to calculate the focal length and power of the contact lenses he requires. Using the lens formula, we determine that Harry needs contact lenses with a focal length of 1.2 meters. The power of these lenses is approximately 0.8333 diopters.

Step by step solution

01

Understand the lens formula

The lens formula relates the object distance (u), image distance (v) and the focal length (f) of a lens. The formula is given as: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \) In this exercise, we are given the near point distance (u) and asked to find the focal length (f) while keeping the image distance (v) at the standard near point distance of 25 cm (0.25 meters).
02

Apply the lens formula

To find the focal length of the contact lenses, we need to substitute the given values of u and v into the lens formula: \( \frac{1}{f} = \frac{1}{1.5} + \frac{1}{0.25} \)
03

Calculate the focal length

Solve the lens formula for f: \( \frac{1}{f} = \frac{1}{1.5} + \frac{1}{0.25} = \frac{1}{1.5} + \frac{4}{1} = \frac{5}{6} \) \( f = \frac{6}{5} = 1.2 \ \mathrm{m} \) Thus, Harry requires contact lenses with a focal length of 1.2 meters.
04

Calculate the power of the lenses in diopters

The power (P) of a lens in diopters is defined as the reciprocal of its focal length in meters: \( P = \frac{1}{f} \) Substitute the value of f obtained in step 3: \( P = \frac{1}{1.2} = 0.8333 \) (rounded to 4 decimal places) Thus, the power of the contact lenses Harry requires is approximately 0.8333 diopters.

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Most popular questions from this chapter

Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?
An object is placed \(12.0 \mathrm{cm}\) in front of a lens of focal length $5.0 \mathrm{cm} .\( Another lens of focal length \)4.0 \mathrm{cm}\( is placed \)2.0 \mathrm{cm}$ past the first lens. (a) Where is the final image? Is it real or virtual? (b) What is the overall magnification? (interactive: virtual optics lab).
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