Problem 62 - College Physics

Question

Water drop magnifier. You can make a pretty good magnifying lens by putting a small drop of water on a piece of transparent kitchen wrap. Suppose your drop has an upper surface with a radius of curvature of 1.6 \(\mathrm{cm}\) and the side on the kitchen wrap is essentially flat. (a) Calculate the focal length of your water lens. (b) What's the angular magnification of the lens? (c) Suppose you place this planoconvex water lens directly onto the surface of a table, so that the tabletop is in effect about half the thickness of the drop. or 1.0 \(\mathrm{mm}_{\text { a away }}\) from the lens. Where does the image of the tabletop form, what type is it, and what is its magnification? (Use the thin lens equation here, even though the small object distance relative to the thickness of the lens makes it a poor approximation in this case.) What does this result tell you about how a simple magnifier works?

Step-by-Step Solution

Answer

(a) Focal length: 4.85 cm; (b) Angular magnification: 6.15; (c) Virtual image 0.104 cm below lens, magnification: 1.04.

Step 1: Understanding Lens Curvature

The problem involves a planoconvex lens, where one side is flat and the other is curved. The radius of curvature given for the curved side is 1.6 cm.

Step 2: Calculate Focal Length

For a lens with one flat side and one curved side, the focal length can be approximated using the lens maker's equation. The equation is \( \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) \), where \( n \) is the refractive index of water (about 1.33) and \( R \) is the radius of curvature. This simplifies to \( \frac{1}{f} = (1.33 - 1) \left( \frac{1}{1.6} \right) \), leading to \( f \approx 4.85 \text{ cm} \).

Step 3: Determine Angular Magnification

Angular magnification \(M_a\) for a simple magnifier is given by the formula \( M_a = \frac{D}{f} + 1 \), where \(D\) is the near point distance, typically 25 cm. Substituting \(f = 4.85\;\mathrm{cm}\), \( M_a \approx \frac{25}{4.85} + 1 \approx 6.15 \).

Step 4: Calculate Image Position

Place the lens 1.0 mm above the table. Using the thin lens equation \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), with \(d_o = 0.1\) cm and \(f = 4.85\) cm. Solving for \(d_i\):\[ \frac{1}{4.85} = \frac{1}{0.1} + \frac{1}{d_i} \Rightarrow \frac{1}{d_i} = \frac{1}{4.85} - 10. \] \(d_i\) turns out to be negative, indicating a virtual image. Calculate to get \(d_i \approx -0.104\;\mathrm{cm}\).

Step 5: Assess Image Type and Magnification

The image is virtual and located approximately -0.104 cm on the same side as the object, signifying it is upright and magnified. Magnification is given by \( M = -\frac{d_i}{d_o} \), calculating \(M \approx 1.04\).

Step 6: Interpretation of the Result

A simple magnifier produces a magnified virtual image. The accuracy of using the thin lens equation here is limited by the small lens thickness, but it shows how magnifiers enlarge objects by forming virtual images on the same side.

Key Concepts

Lens Curvature

When discussing lenses, curvature refers to how curved the surfaces of the lens are. This property significantly influences how light rays are bent when passing through the lens. In this exercise, we are dealing with a planoconvex lens, which means one side is flat and the other is curved with a given radius. The curvature of the lens's surface affects the focal length since it determines how sharply the lens can focus light rays. For our water drop, the radius of curvature is given as 1.6 cm. A smaller radius typically means the lens has a stronger curvature, which can lead to a shorter focal length, enhancing the magnifying power of the lens. Understanding the curvature radius helps in comprehending how effectively the lens can converge or diverge light, depending on its shape.

Focal Length Calculation

The focal length of a lens is the distance from the lens at which parallel incoming light rays are focused to a point. For a lens with one flat and one curved surface, like a planoconvex lens, the focal length is calculated using the lens maker's equation:

• \( \frac{1}{f} = (n - 1) \left( \frac{1}{R} \right) \)
• Here, \(n\) is the refractive index of water, which is 1.33, and \(R\) is the radius of curvature, 1.6 cm.
• Substitute the values into the equation to find \( f \approx 4.85 \text{ cm} \).

This calculation tells us how far from the lens a sharp image will form. A short focal length implies the lens can magnify objects significantly when used as a magnifying lens. It is essential to remember that this is an approximation that hones in on the convergence of light to understand the lens's power.

Angular Magnification

Angular magnification shows how much larger an object appears when viewed through a lens compared to the naked eye. For a simple magnifier, the formula is:

• \( M_a = \frac{D}{f} + 1 \)
• \(D\) is the near point distance, usually 25 cm for a healthy human eye, and \(f\) is the focal length we calculated.
• Plugging in the values: \( M_a \approx \frac{25}{4.85} + 1 \approx 6.15 \).

This indicates a substantial magnification factor, meaning the object appears over six times larger. This is the essence of how magnifiers work; they allow us to see fine details that are too small for the unaided eye by increasing the angular size of the image.

Virtual Image Formation

A virtual image refers to an image formed where the light rays diverge after passing through the lens. Unlike real images, virtual images can't be projected onto a screen. In this context, the water drop forms a virtual image at roughly -0.104 cm when the lens is 1.0 mm above the table. The equation used is the thin lens equation:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
• Given \(d_o = 0.1\) cm (distance from the lens), and using the calculated \(f = 4.85\) cm, we solve for \(d_i\).
• A negative \(d_i\) indicates that the image is virtual and on the same side of the lens as the object, which reveals how virtual images are upright and cannot be captured on paper.

The magnification for this setup shows modest enlargement. Overall, the creation of a virtual image is crucial in understanding how magnifying lenses increase the apparent size of objects we see.