Question

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?

Short answer

The smallest distance \(d\) can have is \(4f\).

Step-by-step solution

Understand the Lens Formula

The lens formula relates object distance (\(u\)) to image distance (\(v\)) and focal length (\(f\)):\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]where \(u + v = d\). We need to find the condition for the smallest \(d\) for which an image is in focus.

Express the Relationship Between Distances

Since the object and image distances add up to \(d\) (i.e., \(u + v = d\)),substitute for \(u\):\[u = d - v\]. Substitute \(u = d - v\) back into the lens formula:\[\frac{1}{f} = \frac{1}{d - v} + \frac{1}{v}\]and simplify to get the condition for \(d\).

Solve for the Image Distance

From the previous step:\[\frac{1}{f} = \frac{1}{d-v} + \frac{1}{v} = \frac{v + (d - v)}{v(d-v)}\]Simplifying,\[\frac{1}{f} = \frac{d}{v(d-v)}\]Rearranging, we find:\[fd = v(d-v)\].

Determine Condition for Minimum d

Substitute \(v = m\) (the image distance) and rearrange:\[fd = m(d - m)\]\[f = \frac{md - m^2}{d}\]The goal is to minimize \(d\). Recognize the condition for the minimum involves the derivative and \(d \geq 4f\). Hence, \(m = u = v\) when \(d\) is minimized for an image in focus.

Calculate the Minimum Distance

Finding minimum \(d\), substitute \(v = d/2\) (since it gives the smallest product for a given sum):\[f = \left( \frac{d/2 \cdot d/2}{d} \right)\]Therefore, \[d = 4f\].

Key concepts

Converging Lens

A converging lens is a transparent optical device that bends light rays towards each other as they pass through it. This type of lens is thicker at the center than at the edges. Converging lenses are crucial in various optical applications like cameras and eyeglasses because they can focus light to form clear images.

When parallel rays of light hit a converging lens, they get refracted in such a way that they meet at a point called the focus. This behavior is highly useful in focusing images on specific planes, such as a screen or sensor, which we often encounter in practical scenarios.

Focal Length

The focal length of a lens is the distance from the lens to the focus, where the refracted light rays meet. The value of the focal length determines how strongly the lens converges or diverges light.

For a converging lens, the focal length is positive and is an essential parameter in the lens formula: \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]where \(f\) is the focal length. The focal length directly impacts where the image appears in relation to the lens, whether it's close or far on the other side of the lens.

Object Distance

The object distance is the distance from the object to the lens. It's a critical component in the lens formula as it helps determine the location of the formed image.

Given the equation:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]we can see that object distance, represented by \(u\), plays a fundamental role in whether the image will form on the screen or not. In practical scenarios, altering the object distance while keeping the focal length constant can help bring the image into proper focus.

Image Distance

Image distance refers to the distance between the lens and the image it forms. Like object distance, it is a crucial part of the lens formula.

• When the image distance \(v\) is positive, it indicates a real image formed on the opposite side of the object.
• Real images can be projected onto a screen, which is common when using converging lenses.

The calculation of the image distance becomes particularly important in determining how far the screen should be placed, allowing for clear, focused images. In the solution above, rearranging the lens formula provides insights into achieving the minimum required distance \(d\), which is key to ensuring the image is in focus on the screen.