Problem 121 - Fundamentals of Physics

Question

An object is \(20 \mathrm{~cm}\) to the left of a thin diverging lens that has a \(30 \mathrm{~cm}\) focal length. (a) What is the image distance \(i ?\) (b) Draw a ray diagram showing the image position.

Step-by-Step Solution

Answer

Image distance \( i = 60 \text{ cm} \). The image is virtual and on the object side.

Step 1: Understanding the Given Information

We are dealing with a thin diverging lens. The object distance (from the lens) is \( o = -20 \text{ cm} \). The negative sign indicates that the object is placed on the opposite side from which the light is incident. The focal length of the lens is \( f = -30 \text{ cm} \). Again, the negative sign is used for diverging lenses.

Step 2: Applying the Lens Formula

To find the image distance \( i \), we use the lens formula: \[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]Substitute the given values, \( f = -30 \text{ cm} \) and \( o = -20 \text{ cm} \). \[\frac{1}{-30} = \frac{1}{-20} + \frac{1}{i}\]

Step 3: Solving the Equation for Image Distance

Rearrange the equation:\[\frac{1}{i} = \frac{1}{-30} - \frac{1}{-20}\]Calculate the right side:\[\frac{1}{i} = \left( \frac{-1}{30} + \frac{1}{20} \right)\]Finding a common denominator (60), solve:\[\frac{1}{i} = \left( \frac{-2}{60} + \frac{3}{60} \right) = \frac{1}{60}\]Therefore, \( i = 60 \text{ cm} \). This is a positive value indicating the image is virtual and formed on the same side as the object.

Step 4: Drafting the Ray Diagram

To draw the ray diagram for a diverging lens:1. Draw a lens and place the principal axis.2. Mark the focal points at \( 30 \text{ cm} \) on both sides (since focal length is 30 cm).3. Place an object arrow \( 20 \text{ cm} \) from the lens on the left side.4. Draw a ray parallel to the principal axis from the top of the object to the lens and then refract it away from the lens diverging upward.5. Draw a second ray through the center of the lens which goes straight without bending.6. Extend the diverging refracted ray backward to meet the above ray.7. The intersection gives the virtual image position, provided at \( 60 \text{ cm} \) on the object side of the lens.

Key Concepts

Lens Formula

The lens formula forms the foundation for understanding how lenses, like diverging lenses, function when it comes to image creation. This formula is expressed as: \[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\] Here, \(f\) is the focal length of the lens, \(o\) is the object distance, and \(i\) represents the image distance.
The diverging lens in our exercise has a focal length of \(-30\) cm, with the negative sign indicating its diverging nature.
When we place our object \(20\) cm from the lens on the left side (hence, \(-20\) cm), the lens formula allows us to calculate where the image will form by rearranging it to solve for \(i\).
After substituting the given values and solving the equation, we find that the image distance \(i\) is \(60\) cm. This positive result indicates something special about the image, which we'll explore next.

Virtual Image

In the world of optics, an image can either be real or virtual, and the calculation from the lens formula helps tell the difference.
A virtual image is formed on the same side as the object, resulting in a positive image distance when calculated with a diverging lens.
Our example demonstrates this perfectly, with the image forming at \(60\) cm on the same side of the lens as the object.Key characteristics of a virtual image include:

• It cannot be projected onto a screen, as it is formed by the apparent divergence of rays.
• It is usually upright and smaller in size compared to the actual object.

Diverging lenses are unique because they always form virtual images. Since the rays after passing through the diverging lens never actually converge, the perceived meeting point of the backward extensions of these rays forms the virtual image.
Understanding virtual images helps in recognizing how lenses affect light paths and create images.

Ray Diagram

Drawing a ray diagram is a practical way to visually understand how a lens, like a diverging lens, forms images.
In our exercise, we use two main rays to determine where the virtual image is located:

• The first ray travels parallel to the principal axis from the top of the object, gets refracted at the lens, and diverges away, seemingly from the focal point on the object side.
• The second ray goes through the center of the lens without changing its path.

Beyond the lens, the divergent path of the first ray is extended backward until it intersects with the second ray.
This intersection point indicates where the virtual image is formed.
Since the rays seem to originate from this point, the virtual image behaves as if it were located there, providing insight into how the image forms with respect to the lens and object. Constructing ray diagrams for diverging lenses not only helps visualize the path of light rays but also confirms the mathematical findings of the lens formula.
It is a useful technique that merges theory with visual intuition.