Problem 78 - Algebra and Trigonometry

Question

The Lens Equation If \(F\) is the focal length of a convex lens and an object is placed at a distance \(x\) from the lens, then its image will be at a distance \(y\) from the lens, where \(F, x,\) and \(y\) are related by the lens equation $$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$$ Suppose that a lens has a focal length of 4.8 \(\mathrm{cm}\) and that the image of an object is 4 \(\mathrm{cm}\) closer to the lens than the object itself. How from the lens is the object?

Step-by-Step Solution

Answer

The object is approximately 11.07 cm from the lens.

Step 1: Understand the Given Information

We are given the focal length of the lens, \( F = 4.8 \, \text{cm} \). We know the image is 4 cm closer to the lens than the object, which gives us the relation \( y = x - 4 \). We need to find the value of \( x \), which is the distance of the object from the lens.

Step 2: Substitute the Relation into the Lens Equation

We start with the lens equation: \[ \frac{1}{F} = \frac{1}{x} + \frac{1}{y} \]Substituting the known values and relationships into the equation, we get:\[ \frac{1}{4.8} = \frac{1}{x} + \frac{1}{x-4} \]

Step 3: Simplify the Equation

Solve the equation to a common denominator. The common denominator of \( x \) and \( x - 4 \) is \( x(x-4) \). Thus, the equation becomes:\[ \frac{x-4 + x}{x(x-4)} = \frac{1}{4.8} \] This simplifies to:\[ \frac{2x - 4}{x(x-4)} = \frac{1}{4.8} \]

Step 4: Cross Multiply to Remove the Denominator

Cross multiply to solve for \( x \):\[ 4.8(2x - 4) = x(x - 4) \]Distribute and bring all terms to one side to form a quadratic equation:\[ 4.8 \cdot 2x - 19.2 = x^2 - 4x \]

Step 5: Simplify and Solve the Quadratic Equation

Combine like terms and rearrange the equation:\[ x^2 - 12.8x + 19.2 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -12.8 \), and \( c = 19.2 \). This gives:\[ x = \frac{12.8 \pm \sqrt{12.8^2 - 4 \cdot 19.2}}{2} \]

Step 6: Calculate with the Quadratic Formula

Calculate the discriminant:\[ 12.8^2 - 4 \cdot 19.2 = 163.84 - 76.8 = 87.04 \]Find the roots using \( x = \frac{12.8 \pm \sqrt{87.04}}{2} \). Calculate this to find:\[ x = \frac{12.8 \pm 9.332}{2} \]

Step 7: Evaluate the Possible Solutions

The two potential solutions for \( x \) are:\[ x = \frac{12.8 + 9.332}{2} = 11.066 \, \text{cm} \] and\[ x = \frac{12.8 - 9.332}{2} = 1.734 \, \text{cm} \]Select the appropriate solution based on the context, where object distance is typically greater.

Key Concepts

Focal Length

Focal length is a crucial concept in understanding how lenses work. It's the distance between the lens and its focus, the point where parallel rays of light converge.
For a convex lens, which bulges outward, the focal length is positive. This type of lens converges light rays to a point, creating a real image if an object is placed outside its focal length.
In our exercise, the focal length is given as 4.8 cm. This is a common parameter for lenses used in optical devices like cameras or glasses. Knowing the focal length allows us to predict how the lens will affect light and create images. We use it in the lens equation to calculate where an image of an object will form when the object is at a certain distance from the lens.
Understanding focal length helps in various applications such as photography, where lenses with different focal lengths are used for capturing wide-angle or zoomed images.

Convex Lens

A convex lens is a transparent optical element that bends light rays inward, causing them to converge at a point. This converging ability makes convex lenses essential in a wide range of optical devices.
Convex lenses are thicker in the middle and thinner at the edges, which distinguishes them from concave lenses that diverge light. The converging property is why convex lenses can form real or virtual images depending on the position of the object relative to the focal point.

• Real images are formed when the object is outside the focal length and appear on the opposite side of the lens.
• Virtual images form when the object is inside the focal length and appear on the same side as the object.

Our problem involved a convex lens with a focal length of 4.8 cm. By using the lens equation, we determined how far the object was from the lens, which is a key aspect of using lenses in real-life scenarios.

Quadratic Equation

Quadratic equations often come into play when dealing with lenses due to the non-linear relationships they describe. These equations are in the form \( ax^2 + bx + c = 0 \) and are crucial for finding solutions that describe positioning in optical setups.
In the lens problem, substituting the relationships into the lens equation and simplifying led to the quadratic equation \( x^2 - 12.8x + 19.2 = 0 \). Solving this equation helps determine the possible positions the object can occupy.

• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) gives us two potential solutions, representing two possible distances the object could be from the lens.
• After calculating, the roots were approximately 11.066 cm and 1.734 cm.

The context, such as which distance is more feasible for a real-world scenario, helps select the correct solution. Thus, quadratic equations are not just about finding values but also understanding their implications in physical situations.