Problem 41 - College Physics

Question

\(\cdot\) An insect 3.75 \(\mathrm{mm}\) tall is placed 22.5 \(\mathrm{cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{cm},\) and the index of refraction of the lens material is 1.70 . (a) Calcu- late the location and size of the image this lens forms of the insect. Is it real or virtual? erect or inverted? (b) Repeat part (a) if the lens is reversed.

Step-by-Step Solution

Answer

Image is virtual, erect, and magnified (17.7 mm high). Reversing lens doesn't change this.

Step 1: Understand the Lens Formula

The lens formula relates the object distance (\(u\)), the image distance (\(v\)), and the focal length (\(f\)) of the lens: \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]. We can use this formula to find the image distance.

Step 2: Calculate the Focal Length

For a planoconvex lens, the focal length \(f\) is determined by the lens maker's formula: \[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]. Here, \(n = 1.70\), \(R_1 = \infty\) (flat surface), and \(R_2 = 13.0\, \mathrm{cm}\). Substitute these values: \[\frac{1}{f} = (1.70-1) \left( \frac{1}{\infty} - \frac{1}{13.0} \right)\]. \[\frac{1}{f} = 0.70 \left( 0 - \frac{1}{13.0} \right)\]. \[\frac{1}{f} = - \frac{0.70}{13.0}\]. The focal length \(f = -18.57 \mathrm{cm}\).

Step 3: Use the Lens Formula to Find Image Distance

Substitute \(u = -22.5\,\mathrm{cm}\) and \(f = -18.57\,\mathrm{cm}\) into the lens formula: \[\frac{1}{-18.57} = \frac{1}{v} - \frac{1}{22.5}\]. Solve for \(v\): \[\frac{1}{v} = \frac{1}{-18.57} + \frac{1}{22.5}\]. Calculate \(v\): \[\frac{1}{v} = -0.0538 + 0.0444\]. \[\frac{1}{v} = -0.00941\]. \(v \approx -106.3\, \mathrm{cm}\).

Step 4: Determine Image Characteristics

The negative \(v\) means the image is virtual and located on the same side as the object. To determine if the image is inverted or erect, we use magnification: \(M = \text{magnification} = \frac{v}{u}\), which also equals the ratio of image height to object height. Calculate magnification: \(M = \frac{-106.3}{-22.5} = 4.72\). Since \(M > 0\) and greater than 1, the image is erect and magnified. The image height \(h' = M \times 3.75\,\mathrm{mm} = 4.72 \times 3.75\,\mathrm{mm} \approx 17.7\,\mathrm{mm}\).

Step 5: Reverse the Lens and Calculate Again

Reversing the lens changes \(R_1 = -13.0\, \mathrm{cm}\) and \(R_2 = \infty\). Use lens maker's formula with these new values: \[\frac{1}{f} = (1.70-1) \left(\frac{1}{-13.0} - \frac{1}{\infty} \right)\]. \[\frac{1}{f} = -\frac{0.70}{13.0}\] as before, resulting in \(f = -18.57 \mathrm{cm}\), indicating the result is unchanged when reversed.Use lens formula for the reversed lens and verify image properties remain the same:For both setups, a negative focal length confirmed the lens configuration's virtual image feature.

Key Concepts

Lens Formula

The lens formula is a core element in geometric optics used to relate the distances involved in image formation. Specifically, it connects the object distance (\(u\)), image distance (\(v\)), and the lens's focal length (\(f\)) through the equation: \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}.\]To solve problems in optics, we rearrange this formula based on known values to find the unknown variable. This is particularly helpful in determining where the image will form for a given object placement. The key is remembering:

• If \(u\) and \(v\) are negative, the object or image is on the same side as the incoming light.
• If \(f\) is negative, the lens is diverging, pointing towards a virtual focus.

Understanding this formula allows you to predict the behavior of optical systems with precision.

Focal Length

The focal length of a lens is critical in determining how it converges or diverges light. For a planoconvex lens, the focal length is derived from the lens maker's formula:\[\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right),\]where \(n\) is the refractive index of the lens material, and \(R_1\) and \(R_2\) are the radii of curvature for the lens surfaces. In this exercise:

• The flat side signifies \(R_1 = \infty\), meaning its contribution is negligible.
• \(R_2\) has a given value affecting the focal length determination.

A negative focal length in our example indicates a virtual focus, established due to the lens's shape and material properties. Recognizing how to use and interpret these elements is crucial for accurate calculations.

Image Characteristics

Image characteristics—such as whether an image is real or virtual, erect or inverted—are largely determined through calculations involving lens variables. The image distance (\(v\)) obtained from the lens formula informs whether an image is real or virtual:

• Negative \(v\) suggests a virtual image formed on the same side as the object.
• Positive \(v\) indicates a real image formed on the opposite side.

The magnification (\(M\)) gives insight into image size and orientation, computed as:\[M = \frac{v}{u},\]and also equals the image height divided by the object height. An erect image results in a positive magnification, revealing upright characteristics compared to the object. Similarly, a magnification greater than 1 means the image is larger than the original object, offering a quick judgment on size changes.

Planoconvex Lens

A planoconvex lens is a unique optical element with one flat (plano) and one convex surface. This design helps in controlling the convergence of light:

• The plano side ensures minimal refraction and aids in directing light through the lens.
• The convex side focuses or diverges light depending on its curvature.

Such lenses are common in applications requiring focus control and light manipulation, benefiting from simple yet effective geometric curvature changes. By reversing the lens as in our exercise, the same lens properties apply; however, the curvature logic shifts with the adjusted radius. Understanding planoconvex lenses offers clarity in their use, from cameras to corrective eyewear, highlighting their versatility in focusing and redirecting optical paths.