Some electro-optic materials can change their index of refraction in response to an applied voltage. Suppose a plano-convex lens (flat on one side, a 15.0 cm radius of curvature on the other), made from a material whose normal index of refraction is 1.500, is creating an image of an object that is 50.0 cm from the lens. By how much would the index of refraction need to be increased to move the image 5.0 cm closer to the lens?

Verified step by step guidance

1. Step 1: Begin by understanding the lens maker's equation, which relates the focal length of a lens to its radius of curvature and index of refraction. The equation is: $\frac{1}{f}=(n-1)\left[\frac{1}{R}\right]$, where $f$ is the focal length, $n$ is the index of refraction, and $R$ is the radius of curvature.
2. Step 2: Use the thin lens equation to relate the object distance ($d$_o), image distance ($d$_i), and focal length ($f$): $\frac{1}{f}=\frac{1}{d}$_o+1d_i. Substitute the given object distance ($50.0$ cm) and the initial image distance ($55.0$ cm, since the image is moved 5.0 cm closer). Solve for the initial focal length.
3. Step 3: Recalculate the focal length after the image is moved 5.0 cm closer to the lens. Use the thin lens equation again, but this time with the new image distance ($50.0$ cm). This will give the new focal length required to achieve the new image position.
4. Step 4: Use the lens maker's equation to find the new index of refraction ($n$) that corresponds to the new focal length. Rearrange the lens maker's equation to solve for $n$: $n=1+\frac{f}{R}$. Substitute the new focal length and the radius of curvature ($15.0$ cm).
5. Step 5: Calculate the change in the index of refraction by subtracting the original index of refraction ($1.500$) from the new index of refraction obtained in Step 4. This will give the required increase in the index of refraction to move the image 5.0 cm closer to the lens.

Key Concepts

Index of Refraction

The index of refraction is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. A higher index indicates that light travels slower in that medium, which affects how light bends when entering or exiting the material.

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens, expressed as 1/f = 1/v + 1/u. This formula is crucial for determining the position of the image formed by a lens based on the object's position and the lens's characteristics, including its curvature and refractive index.

Electro-Optic Effect

The electro-optic effect refers to the change in the optical properties of a material in response to an applied electric field. In the context of electro-optic materials, this effect can alter the index of refraction, allowing for dynamic control of light propagation, which is essential for applications like modulators and switches in optical systems.