Problem 94 - Intermediate Algebra

Question

The focal length, \(f,\) of a lens is given by the lensmaker's formula, $$\frac{1}{f}=0.6\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right)$$ where \(f\) is the focal length of the lens and \(r_{1}\) and \(r_{2}\) are the radii of the two circular surfaces. Solve the formula for \(f\).

Step-by-Step Solution

Answer

The focal length \( f \) is \( f = \frac{1}{0.6} \left( \frac{1}{r_1} + \frac{1}{r_2} \right)^{-1} \).

Step 1: Understand the Lensmaker's Formula

The given formula is \( \frac{1}{f} = 0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \). This formula relates the focal length \( f \) of a lens to the radii \( r_1 \) and \( r_2 \) of the curved surfaces of the lens.

Step 2: Isolate \( \frac{1}{f} \)

Notice that the formula is already solved for \( \frac{1}{f} \). It states that \( \frac{1}{f} \) is equal to \( 0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \). We need to invert this to solve for \( f \).

Step 3: Invert the Equation

To isolate \( f \), you need to take the reciprocal of both sides. This will give you \( f = \frac{1}{0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right)} \).

Step 4: Simplify the Expression

The simplified expression for \( f \) becomes \( f = \frac{1}{0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right)} \). Alternatively, you can express this as \( f = \frac{1}{0.6} \times \frac{1}{ \left( \frac{1}{r_1} + \frac{1}{r_2} \right) } \).

Key Concepts

Focal Length

The focal length of a lens, often denoted by \(f\), is a crucial concept in optics. It represents the distance from the lens to the point where parallel rays of light converge or appear to diverge. This focal point plays a key role in determining how a lens magnifies or focuses an image. A shorter focal length means the lens has a higher optical power and can magnify objects to a greater degree. Conversely, a longer focal length results in a lower magnification.
Understanding the focal length helps in designing lenses for different applications, whether it is a camera lens, microscope, or glasses. The lensmaker's formula, as seen in the exercise, is instrumental in calculating the exact focal length based on the lens's shape and material properties.

Radius of Curvature

The radius of curvature, denoted by \( r_1 \) and \( r_2 \) for a lens with two surfaces, tells us how curved the lens surfaces are. Each surface of a lens can be thought of as a part of a sphere, and the radius of this sphere is what we call the radius of curvature.

• If \( r \) is large, the surface is relatively flat, like the earth from horizon to horizon.
• If \( r \) is small, the surface is significantly curved, much like a bowl or dome.

These radii impact the lens’s ability to bend light. The lensmaker's formula uses these values to help calculate the focal length, as more pronounced curves can bring light to a focus more quickly than a flatter piece.

Algebraic Manipulation

Algebraic manipulation is key to solving mathematical problems, such as the given exercise involving the lensmaker's formula. It involves rearranging equations to isolate the desired variable, in this case, the focal length \( f \).
Initially, the goal was to take the given expression \( \frac{1}{f} = 0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \) and manipulate it to solve for \( f \). This required:

• Recognizing that \( \frac{1}{f} \) is already isolated on one side of the equation.
• Inverting the entire equation by taking the reciprocal of both sides to solve for \( f \). This gives us \( f = \frac{1}{0.6 \left( \frac{1}{r_1} + \frac{1}{r_2} \right)} \).
• Rewriting this in a more intuitive form, such as \( f = \frac{1}{0.6} \times \frac{1}{ \left( \frac{1}{r_1} + \frac{1}{r_2} \right) } \).

Such steps enhance our understanding of the mathematical relationships within the problem, allowing us to accurately find the solutions we need.