Problem 18 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A bright object is placed on one side of a converging lens of focal length \(f,\) and a white screen for viewing the image is on the opposite side. The distance \(d_{\mathrm{T}}=d_{\mathrm{i}}+d_{\mathrm{o}}\) between the object and the screen is kept fixed, but the lens can be moved. \((a)\) Show that if \(d_{\mathrm{T}}>4 f,\) there will be two positions where the lens can be placed and a sharp image will be produced on the screen. (b) If \(d_{\mathrm{T}}<4 f,\) show that there will be no lens position where a sharp image is formed. (c) Determine a formula for the distance between the two lens positions in part \((a),\) and the ratio of the image sizes.

Step-by-Step Solution

Answer

For (a), two lens positions exist if \(d_T > 4f\). For (b), no position exists if \(d_T < 4f\). For (c), derive position distance using quadratic roots and size ratio with magnifications.

Step 1: Understanding the Lens Formula

The lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance.

Step 2: Relate with Total Distance

Given that the total distance \(d_T = d_i + d_o\) is fixed, express the image distance in terms of \(d_T\) and \(d_o\): \(d_i = d_T - d_o\). Substitute this in the lens formula.

Step 3: Derive the Quadratic Equation

By substituting \(d_i = d_T - d_o\) in \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_T - d_o}\), we get a quadratic equation in \(d_o\). Simplify: \(f(d_T - d_o) + f d_o = d_o(d_T - d_o)\).

Step 4: Solve the Quadratic Equation

Rearrange the terms to form a quadratic equation: \[d_o^2 - d_T d_o + f d_T - fd_o = 0\]. Solving for \(d_o\) using the quadratic formula \(d_o = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the two roots for \(d_o\).

Step 5: Condition for Two Real Solutions

For two real and distinct solutions, the discriminant of the quadratic equation must be positive: \((d_T)^2 - 4f d_T > 0\). Simplify: \(d_T > 4f\).

Step 6: No Real Solutions Case

If \(d_T < 4f\), then the discriminant \((d_T)^2 - 4f d_T < 0\), meaning no real solutions exist, hence no sharp images can be formed.

Step 7: Calculate Distance Between Lens Positions

The solutions of the quadratic equation give the lens positions. Calculate the distance between these positions by finding the difference of the roots.

Step 8: Image Sizes Ratio

Use magnification formula \(m = -\frac{d_i}{d_o}\) to find the image sizes for both positions. Determine the ratio of magnifications as \(\frac{m_1}{m_2} = \left(\frac{-d_{i1}}{d_{o1}}\right) \left(\frac{d_{o2}}{-d_{i2}}\right)\).

Key Concepts

Converging Lens

A converging lens, often called a convex lens, is a tool in optics that helps focus light. This type of lens is thicker at the center than at the edges.
This shape allows it to bend incoming parallel light rays to a focal point on the other side of the lens.

• The focal point is where the light converges. If you place an object at the focal point, it generates a parallel beam of light.
• Converging lenses can create real images—these images can be captured on a screen, as they actually form on the other side of the lens.

Understanding the nature of a converging lens is essential for comprehending how it can manipulate light to form images under varying conditions.

Lens Formula

In optics, the lens formula mathematically relates the focal length (\(f\)), object distance (\(d_o\)), and image distance (\(d_i\)). This relationship is expressed as:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
This equation helps determine the position where an image will form and is foundational for solving many lens problems.

• If you know any two variables among \(f\), \(d_o\), and \(d_i\), you can find the third.
• This formula is crucial when working with fixed distances and moving the lens to find different image positions, as in the provided exercise.

Grasping this concept allows deeper insights into how lenses create images by altering distances.

Image Formation

Image formation involves determining where an image will appear relative to a lens and understanding its nature.
For a converging lens, when an object is placed beyond the focal length, a real and inverted image forms on the opposite side. This is where it can be captured on a screen.
Factors affecting image formation include:

• Object Distance (\(d_o\)) - The placement of the object relative to the lens alters image characteristics.
• Focal Length (\(f\)) - Determines the confluence of light and thus affects image clarity and size.
• Lens Position - Adjusting the lens position helps in finding the spot where the image becomes sharp.

Successful image formation requires balancing these elements precisely.

Quadratic Equation

Quadratic equations often emerge in optics when dealing with lenses, especially when solving problems with distances kept constant.
From the lens formula, a quadratic equation can stem when rearranging terms for particular values, such as total distance \(d_T = d_i + d_o\).
The exercise leads to this form: \[d_o^2 - d_T d_o + f d_T - fd_o = 0\]
This equation yields two possible solutions (or roots) using the quadratic formula:\[d_o = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• If the discriminant (\(b^2 - 4ac\)) is positive, there are two real and distinct roots, indicating that two lens positions exist for sharp images.
• A negative discriminant implies no real solutions, and thus no sharp image formation.

Understanding this quadratic nature is key to predicting image formation scenarios.

Magnification

Magnification describes how much larger or smaller an image is compared to the object.
It is expressed in terms of object and image distances as:
\[m = -\frac{d_i}{d_o}\]
Here:

• The negative sign indicates that the image formed by a converging lens is inverted.
• Understanding different lens positions is crucial, as it affects the image distance \(d_i\), thereby changing magnification.

The ratio of magnifications for two lens positions provides insight into their relative image sizes:\[\frac{m_1}{m_2} = \left(\frac{-d_{i1}}{d_{o1}}\right) \left(\frac{d_{o2}}{-d_{i2}}\right)\]
Analyzing such ratios helps determine different characteristics of images formed under varied spatial configurations.