Question

A lens of a pair of eyeglasses has a power of \(5.55 \mathrm{D}\) in air. The power of this same lens is 2.262 if it is put in water with \(n=1.333 .\) What is the index of refraction of this lens?

Short answer

Based on the step-by-step solution, create a short answer question: Question: A lens has a power of 5.55 D in air and 2.262 D in water. The index of refraction of water is 1.333. Find the index of refraction of the lens. Answer: To find the index of refraction of the lens, first use the lensmaker's equation to write two equations for the lens in air and water. Next, solve the equations for the focal length f. Finally, substitute the value of f into either equation for n_lens and calculate the index of refraction for the lens. In this case, the index of refraction for the lens is approximately 1.49.

Step-by-step solution

Write down the known values.

We are given: Power of the lens in air (P1) = 5.55 D Power of the lens in water (P2) = 2.262 D Index of refraction of water (n_water) = 1.333 Our goal is to find the index of refraction of the lens n_lens.

Use the lensmaker's equation.

The lensmaker's equation is: \(P = (n_{lens} - n_{medium})/f\) Where P is the power of the lens, n_lens is the index of refraction of the lens, n_medium is the index of refraction of the surrounding medium, and f is the focal length of the lens. Since we have the power of the lens in both air and water, we can write two equations: 1. In air: \(P_1 = (n_{lens} - n_{air})/f\) 2. In water: \(P_2 = (n_{lens} - n_{water})/f\)

Solve the equations for n_lens.

First, rearrange the equations to represent n_lens in terms of f: 1. In air: \(n_{lens} = P_1f + n_{air}\) 2. In water: \(n_{lens} = P_2f + n_{water}\) Now set the two equations equal and solve for f. \((P_1f + n_{air}) = (P_2f + n_{water})\) \(f = (n_{water}-n_{air})/(P_1 - P_2) = (1.333-1)/(5.55-2.262)\) By calculating the above expression, we find the focal length f.

Calculate the index of refraction for the lens.

Now, we can substitute the value of f we found in Step 3 into either equation for n_lens. In this case, we will use the equation for air: \(n_{lens} = P_1f + n_{air}\) \(n_{lens} = 5.55f + 1\) Insert the value of f obtained in Step 3 and calculate the index of refraction for the lens.

Key concepts

Lens Power Explained

In optics, lens power is a crucial concept. It represents the lens's ability to converge or diverge light. Measured in diopters (D), it is defined as the inverse of the focal length (in meters). So, the power P of a lens is given by the formula:

\[ P = \frac{1}{f} \]
Where f is the focal length. This relationship means that a higher power indicates a shorter focal length and thus, a stronger lens. Corrective eyeglasses often have lenses with specific powers to adjust the user's vision, whether they are nearsighted or farsighted.

The Lensmaker's Equation

The lensmaker's equation is central when understanding how lenses are designed. It relates the physical properties of a lens with its optical characteristics. The general form of the equation for a lens in air is:

\[ P = (n_{lens} - 1)/f \]
In contexts where the lens is immersed in a medium other than air, the equation adjusts to account for the medium's index of refraction (n_medium):

\[ P = (n_{lens} - n_{medium})/f \]
This allows us to calculate the index of refraction of the lens material (n_lens) which indicates how much the material bends light, a fundamental aspect in designing lenses for various applications.

Understanding Focal Length

The focal length of a lens is the distance between the lens and the point where parallel rays of light converge after passing through the lens. It's a direct measure of the lens's optical power — the shorter the focal length, the greater the curvature and the stronger the lens. Focal length is not just a static value; it can change depending on the surrounding medium's index of refraction. As determined by the lensmaker's equation, the focal length is inversely proportional to the lens power and is affected by the relative index of refraction difference between the lens material and the surrounding medium.

Optics in Physics

Optics is the branch of physics that focuses on the study of light behavior and properties, including its interactions with matter. This involves understanding how light is refracted when passing through different media, such as lenses, and how this can be used to correct vision or create magnified images. The lens power, lensmaker's equation, and focal length are all fundamental concepts within optics that help us explain and utilize the refractive properties of lenses in various technologies like glasses, cameras, and telescopes.