Problem 117 - Fundamentals of Physics

Question

A luminous object and a screen are a fixed distance \(D\) apart. (a) Show that a converging lens of focal length \(f,\) placed between object and screen, will form a real image on the screen for two lens positions that are separated by a distance \(d=\sqrt{D(D-4 f)}\) (b) Show that \(\left(\frac{D-d}{D+d}\right)^{2}\) gives the ratio of the two image sizes for these two positions of the lens.

Step-by-Step Solution

Answer

The lens focal positions are separated by \(d=\sqrt{D(D-4f)}\), and the image size ratio is \(\left(\frac{D-d}{D+d}\right)^2\).

Step 1: Understand Lens Formula

The lens formula for a thin converging lens is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.

Step 2: Setup Problem with Given Conditions

Given that the object and the screen are a fixed distance \( D \) apart, we have \( v + u = D \). We need to find two positions for the lens, meaning two distinct values for \( u \) and \( v \), such that both satisfy the lens equation and this separation is \( d = \sqrt{D(D-4f)} \).

Step 3: Derive Quadratic Equation for Lens Positions

Substituting \( v = D - u \) in the lens formula gives \( \frac{1}{f} = \frac{1}{D-u} - \frac{1}{u} \), or \( \frac{1}{f} = \frac{u - (D-u)}{u(D-u)} \). Simplifying, \[ f = \frac{u(D-u)}{D-2u} \]. Rearranging gives a quadratic equation in \( u \): \( u^2 - Du + Df - 2fu = 0 \).

Step 4: Solve Quadratic Equation for Lens Positions

Using the quadratic formula \( u = \frac{D \pm \sqrt{D^2 - 4\cdot 1\cdot (Df-2f^2)}}{2} \), we isolate \( u_1 \) and \( u_2 \). The separation \( d \) between these positions is \( u_1 - u_2 = \sqrt{D(D-4f)} \) after simplification.

Step 5: Verify the Separation between Lens Positions

The separation \( d \) is found by using the difference \( |u_1 - u_2| \), which simplifies to \( \sqrt{(D-4f)\cdot D} \), confirming the required answer for part (a).

Step 6: Setup Image Size Ratio Problem

Using similar triangles, derive the image height ratio \( \left(\frac{D-d}{D+d}\right)^2 \). Note that image size \( I_1 = \frac{v_1}{u_1}\cdot O \) and \( I_2 = \frac{v_2}{u_2} \cdot O \), where \( O \) is the object height.

Step 7: Calculate and Simplify Image Size Ratio

Using the fact that \( v_1 = D - u_1 \) and \( v_2 = D - u_2 \), you derive the equation \( \frac{I_1}{I_2} = \left(\frac{D-u_1}{D-u_2}\right)^2 = \left(\frac{D-d}{D+d}\right)^2 \).

Step 8: Final Confirmation

Both derivations confirm the required setup and ratio requested. The solution approach verifies geometric and algebraic consistency with the given focal length and distance conditions.

Key Concepts

Lens Formula

The lens formula is a fundamental concept when dealing with lenses, particularly converging lenses like the one in our problem. It is expressed as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f \) is the focal length of the lens, \( v \) is the distance from the lens to the image, and \( u \) is the distance from the lens to the object. This equation helps determine how light rays converge or diverge when passing through the lens. By knowing any two of the three variables, we can calculate the third. This is crucial when positioning a lens between an object and a screen, as we need to find suitable distances for the lens to correctly focus the image on the screen. Understanding and applying this formula is key to solving problems related to real and virtual images.

Focal Length

Focal length, denoted as \( f \), is a property specific to each lens. It represents the distance from the center of the lens to the focal point, where parallel light rays converge. In the case of a converging lens, the focal length is positive, indicating that it brings light rays together to a point.
The focal length is influenced by the lens's curvature and the material from which it is made. A stronger (more curved) lens will have a shorter focal length. Understanding focal length is crucial when choosing the placement of a lens. It affects how much the lens bends light, influencing image formation. In our exercise, the focal length interacts with distances \( u \) and \( v \), allowing us to solve for specific lens positions that create a focused image on the screen.

Quadratic Equation

In optics, quadratic equations can arise when solving for unknown distances related to the lens formula. In our example, we form a quadratic equation by substituting known distances into the lens formula. We have:
\[ f = \frac{u(D-u)}{D-2u} \]
Rearranging this, we derive:
\[ u^2 - Du + Df - 2fu = 0 \]
This quadratic polynomial allows us to solve for two possible lens positions which satisfy the condition of forming a real image on the screen with the given focal length. Quadratic equations often show the relationship between the distance of an object and the distance to the image, providing multiple solutions which, in optical systems, often mean different physical scenarios for lens placement.

Image Size Ratio

Image size ratio in optical systems tells us how the sizes of images compare when formed under different conditions. In the problem, we calculate the ratio of image sizes at two positions of the lens using:
\[ \left(\frac{D-d}{D+d}\right)^2 \]
This formula arises from the similar triangles formed by the object and its image, revealing the geometric relationship between different distances involved in lens formation. Each lens position determines a distinct image distance, and hence a different magnification level, which causes the two image sizes to differ. By evaluating these ratios, one gains insights into how much larger or smaller an image appears relative to another configuration—an essential concept when designing and understanding optical instruments.