Problem 60 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. All numbers are accurate to at least two significant digits. For an optical lens, the sum of the reciprocals of \(p,\) the distance of the object from the lens, and \(q\), the distance of the image from the lens, equals the reciprocal of \(f,\) the focal length of the lens. If \(p\) is \(5.0 \mathrm{cm}\) greater than \(q(q>0)\) and \(f=4.0 \mathrm{cm},\) find \(p\) and \(q\).

Step-by-Step Solution

Answer

The distances are \(p = 9 \text{ cm}\) and \(q = 4 \text{ cm}\).

Step 1: Understanding the 1/Lens Formula

The lens formula is given by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(f\) is the focal length. We're provided with \(f = 4.0 \text{ cm}\). We substitute \(f\) into the formula to get \(\frac{1}{4} = \frac{1}{p} + \frac{1}{q}\).

Step 2: Identify Relationship Between p and q

According to the problem, \(p\) is \(5.0 \text{ cm}\) greater than \(q\), i.e., \(p = q + 5.0\). We will use this relationship to substitute \(p\) in the lens formula.

Step 3: Substitute p in the Lens Formula

Replace \(p\) with \(q + 5.0\) in the lens formula: \(\frac{1}{4} = \frac{1}{q + 5.0} + \frac{1}{q}\). This equation will help us solve for \(q\).

Step 4: Clear Fractions and Simplify

Multiply through by \((q + 5.0)q\) to clear the fractions: \[(q + 5.0)q \left(\frac{1}{4}\right) = (q + 5.0)q \left(\frac{1}{q + 5.0} + \frac{1}{q}\right)\]Simplify to get:\[(q + 5.0)q \cdot \frac{1}{4} = q + q(q + 5.0)\]

Step 5: Distribute Terms

On the left side, distribute to obtain \(q^2 + 5q = 4q + 20\). Move all terms to one side to form the quadratic equation: \[q^2 + q - 20 = 0\].

Step 6: Solve the Quadratic Equation

Use the quadratic formula to solve for \(q\): \[q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \(a = 1\), \(b = 1\), and \(c = -20\). Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4\times1\times(-20) = 81\). So, \[q = \frac{-1 \pm \sqrt{81}}{2}\] This simplifies to \[q = \frac{-1 + 9}{2} = 4\] or \[q = \frac{-1 - 9}{2} = -5\]. The negative result is not physical (since \(q > 0\)), so choose \(q = 4 \text{ cm}\).

Step 7: Calculate p Using q

With \(q = 4 \text{ cm}\), substitute back to find \(p\): \[p = q + 5.0 = 4 + 5 = 9 \text{ cm}\].

Key Concepts

Optical Lens Equation

The optical lens equation is a fundamental tool used in field of optics. It helps in finding the relationship between the object distance (p), the image distance (q), and the focal length (f) of a lens. The equation is expressed as:
\[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\]This formula arises from the principle of refraction, which governs how light behaves when it passes through a lens. Using this equation, one can determine either the position of the object, the image, or find the lens focal length, given the other two parameters.
To apply this equation effectively, it's important to keep in mind:

• The distances, p and q, are measured from the lens to the object and image, respectively.
• The positive/negative sign of these distances depends on the lens type and the convention being used.
• Generally, for converging lenses, the focal length is positive, while for diverging lenses, it is negative.

Focal Length Calculation

When tasked with focal length calculation, the optical lens equation can be rearranged to solve for f. Given the values of p and q, finding f involves straightforward algebraic manipulation.
Here's a basic walkthrough:

• If given p and q, you can rearrange the lens formula to \[ f = \frac{pq}{p+q} \].
• Substitute the known values for p and q into the equation to find f.
• Ensure all units are consistent to avoid errors in calculation.

This calculation shows the lens' power to focus light rays and can help in lens design or evaluation. Always check for appropriate units and sign conventions when dealing with real scenarios in optics.

Quadratic Equation in Optics

In the realm of optics, quadratic equations frequently arise, especially when dealing with real-life scenarios like the one given in the original exercise.
When substituting relationships between variables into the lens formula, it might be necessary to handle quadratic expressions. Here's how it often unfolds:

• Through substitution, the lens equation sometimes transforms into a quadratic form like \[ ax^2 + bx + c = 0 \].
• To solve for unknowns such as image or object distance, the quadratic formula is used: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
• Computing the discriminant, b^2 - 4ac, helps determine the nature of roots (real and distinct, real and equal, or complex).

In the context of optics, ensure that the value chosen for variables (such as q in the exercise) is physically meaningful, adhering to the constraints set by the problem like positivity of distances.