Problem 55 - Physics

Question

An object is \(18 \mathrm{~cm}\) in front of a diverging lens that has a focal length of \(-12 \mathrm{~cm} .\) How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of \(2.0 ?\)

Step-by-Step Solution

Answer

Move the object to 12 cm in front of the lens.

Step 1: Understand the Given Information

We have an object placed 18 cm in front of a diverging lens with a focal length of -12 cm. The goal is to adjust the position of the object such that the image size is half of the object's size.

Step 2: Apply the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f = -12 \mathrm{~cm} \) (focal length), and we need to find the new \( u \) (object distance), that results in a magnification of -0.5.

Step 3: Understand Magnification Concept

The magnification \( m \) of a lens is given by \( m = \frac{-v}{u} \). For an image reduced by half the original size, \( m = -0.5 \).

Step 4: Express Relationships between Variables

Using \( m = \frac{-v}{u} = -0.5 \), rearrange to find \( v \) in terms of \( u \): \( v = 0.5u \).

Step 5: Substitute into Lens Formula

Substitute \( v = 0.5u \) into the lens formula: \[ \frac{1}{f} = \frac{1}{0.5u} - \frac{1}{u} \].

Step 6: Simplify the Equation

Simplify \( \frac{1}{-12} = \frac{2}{u} - \frac{1}{u} \) to get \( -\frac{1}{12} = \frac{1}{u} \).

Step 7: Solve for New Object Distance

Invert the equation \( \frac{1}{u} = -\frac{1}{12} \) to find \( u = -12 \mathrm{~cm} \).

Step 8: Interpret Solution

The negative sign indicates that the object should be placed 12 cm in front of the lens (on the same side as the object was initially).

Key Concepts

Lens Formula

The lens formula is a crucial principle when working with lenses, allowing us to determine the relationships between the object distance, image distance, and focal length of a lens. It is given by the equation:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]Here:

• \( f \) is the focal length of the lens. For a diverging lens, the focal length is negative.
• \( v \) represents the image distance from the lens.
• \( u \) denotes the object distance from the lens.

In our exercise, the focal length \( f \) is \(-12 \mathrm{~cm} \). To find how far the object should be placed for specific image characteristics, we utilize this formula.
This equation helps us understand how the variables interact and solve for the unknown when two values are known.

Magnification

Magnification refers to the factor by which the size of an image changes in comparison to the size of the object. It is defined by the formula:\[m = \frac{-v}{u}\]Where:

• \( m \) is the magnification factor. Negative magnification indicates an inverted image.
• \( v \) is the image distance, as measured from the lens.
• \( u \) is the object distance from the lens.

In our context, we're asked to reduce the image size by a factor of 2. This sets our magnification \( m \) to \(-0.5 \), implying an inverted and reduced image. The formula assists in calculating how object distance needs to be adjusted to achieve this magnification.

Focal Length

The focal length of a lens is a measure of how strongly it converges or diverges light. For a diverging lens, like the one in our exercise, it is typically negative, reflecting its property of spreading out light rays. The given focal length here is \(-12 \mathrm{~cm} \).
The focal length influences the image formation significantly, being part of the lens formula:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]This tells us how changes in object distance \( u \) can affect the image distance \( v \). A negative focal length affects the directions in which the calculations are performed, informing us about the virtual and diminished nature of images formed by the diverging lens.

Object Distance

The object distance, represented by \( u \), is the distance from the object to the lens. This is a critical parameter in determining how an image is formed by a lens. In the presented situation, our goal was to find the object distance which results in a magnification of \(-0.5 \).
By manipulating the lens formula and the magnification relation, we determined that:\[\frac{1}{-12} = \frac{1}{v} - \frac{1}{u}\]and substituting for \( v = 0.5u \), we solve to find \( u \). Solving the equation, we found \( u = -12 \mathrm{~cm} \).
The negative result signifies that the object should be positioned 12 cm in front of the lens, confirming that our calculation aligns with the lens formula's conventions for diverging lenses.