Question

An object is placed \(15.0 \mathrm{~cm}\) to the left of a converging lens with focal length \(f=5.00 \mathrm{~cm},\) as shown in the figure. Where is the image formed? a) \(1.25 \mathrm{~cm}\) to the right of the lens b) \(2.75 \mathrm{~cm}\) to the left of the lens c) \(3.75 \mathrm{~cm}\) to the right of the lens d) \(5.00 \mathrm{~cm}\) to the left of the lens e) \(7.50 \mathrm{~cm}\) to the right of the lens

Short answer

Answer: (c) 3.75 cm to the right of the lens. Explanation: Using the lens formula, we find that the image distance (di) is 3.75 cm with a positive sign, which indicates that the image is formed to the right of the lens.

Step-by-step solution

Write down given information

The object is placed at a distance of \(15.0 \mathrm{~cm}\) to the left of the lens, which means the object distance \(d_o = -15.0 \mathrm{~cm}\) (we include the negative sign as the object is to the left of the lens). The focal length of the converging lens is \(f = 5.00 \mathrm{~cm}>0\), by definition. We need to find out where the image is formed, so we need to find its distance from the lens \(d_i\).

Apply the lens formula

Use the lens formula to relate the object distance (\(d_o\)), image distance (\(d_i\)), and focal length (\(f\)): \(\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}\) Plug in the given values for \(d_o\) and \(f\): \(\frac{1}{-15.0} + \frac{1}{d_i} = \frac{1}{5.00}\)

Solve for the image distance

Isolate \(d_i\) in the equation: \(\frac{1}{d_i}=\frac{1}{5.00}+\frac{1}{15.0}\) Now, find the least common denominator for the fractions and add them: \(\frac{1}{d_i}=\frac{3+1}{15}= \frac{4}{15}\) Finally, take the reciprocal of both sides to find \(d_i\): \(d_i=\frac{15}{4}=3.75 \mathrm{~cm}\)

Choose the correct option

Since the calculated \(d_i=3.75 \mathrm{~cm}>0\), the image is formed to the right of the lens. Therefore, the correct answer is (c) \(3.75 \mathrm{~cm}\) to the right of the lens.

Key concepts

Lens Formula

Understanding the lens formula is crucial when analyzing image formation with lenses. This formula provides a relationship between the object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of the lens.

The lens formula is given by the equation \[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]. It is important to remember that distances measured in the direction of incoming light (towards the lens from the object) are considered positive, while distances measured against the direction of incoming light are negative. Focal length \(f\) for converging lenses is positive, and for diverging lenses, it is negative.

The practical use involves plugging in the object distance and the focal length into the lens formula to solve for the image distance, which tells us where the image will be formed relative to the lens.

Object Distance

The object distance, denoted by \(d_o\), is the distance between the object and the lens. It plays a significant role in image formation as part of the lens formula.

In the context of a converging lens, if the object is placed to the left of the lens, \(d_o\) is taken as negative in the equation. This convention stems from the sign convention in optics, where directions are relative to the direction of light propagation assumed to be from left to right.

Understanding the sign and value of \(d_o\), students can better predict the nature of the image (real or virtual) and the configuration of the optics system.

Image Distance

The image distance \(d_i\) refers to the location where the image forms in relation to the lens. Positive \(d_i\) values indicate the image is formed on the opposite side of the lens from the object (real images), while negative values suggest the image is formed on the same side as the object (virtual images).

In our exercise, solving the lens formula for \(d_i\) reveals the position of the image. Once \(d_i\) is found to be positive, as in our solved example at 3.75 cm, this indicates a real image formed on the opposite side of the lens from the object.

Focal Length

Focal length, represented as \(f\), is a measure of how strongly a lens converges or diverges light. It is the distance from the lens to the focal point, the spot where parallel rays of light converge. In a converging lens, \(f\) is positive, and in a diverging lens, it is negative.

In the exercise, the given focal length of 5.00 cm signifies a converging lens. Focal length is integral to determining the image formation, as it affects the bending of light rays and helps in calculating both \(d_o\) and \(d_i\) using the lens formula. A shorter focal length implies stronger bending and vice versa, which influences the size and position of the image produced.