Problem 27 - University Physics with Modern Physics

Question

An insect 3.75 mm tall is placed 22.5 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 cm, and the index of refraction of the lens material is 1.70. (a) Calculate 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

(a) Image is real, inverted, 0.635 m; size 10.6 mm. (b) Same results when lens is reversed.

Step 1: Convert Units

First, convert the distance from centimeters to meters since the standard unit for optical calculations is meters. The insect is at 22.5 cm, which is 0.225 m. The radius of curvature is 13.0 cm, which is 0.13 m.

Step 2: Lens Maker's Equation

Determine the focal length of the lens using the Lens Maker's equation:\[ \frac{1}{f} = (n - 1) \left ( \frac{1}{R_1} - \frac{1}{R_2} \right ) \]For the planoconvex lens, \( R_1 = \infty \) (flat surface) and \( R_2 = -0.13 \text{ m} \). Substitute \( n = 1.70 \):\[ \frac{1}{f} = (1.70 - 1) \left ( \frac{1}{\infty} - \frac{1}{-0.13} \right ) \approx \frac{0.7}{0.13} \]\[ f \approx 0.1857 \text{ m} \]

Step 3: Use Lens Equation

Use the lens equation to find the image distance \( v \):\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]where \( u = -0.225 \text{ m} \) (object is on the left).Substitute the values:\[ \frac{1}{0.1857} = \frac{1}{-0.225} + \frac{1}{v} \]Calculate \( v \):\[ v \approx 0.635 \text{ m} \]

Step 4: Magnification and Image Characteristics

Calculate the magnification \( M \):\[ M = - \frac{v}{u} = - \frac{0.635}{-0.225} \approx 2.82 \]The image height \( h' = Mh \):\( h = 3.75 \text{ mm} = 0.00375 \text{ m} \)\( h' \approx 2.82 \times 0.00375 \approx 0.0106 \text{ m} \text{ or } 10.6 \text{ mm} \)The image is real (since \( v > 0 \)) and inverted (since magnification is negative).

Step 5: Reverse the Lens

Reverse the lens and recalculate. Now \( R_1 = 0.13 \text{ m} \) and \( R_2 = \infty \):\[ \frac{1}{f} = (1.70 - 1) \left ( \frac{1}{0.13} - \frac{1}{\infty} \right ) \approx \frac{0.7}{0.13} \Rightarrow f \approx 0.1857 \text{ m} \]Since the focal length is the same, \( v \) and magnification will also be the same as when lens was not reversed.

Key Concepts

Lens Maker's Equation

The Lens Maker's Equation is a fundamental formula in optics. It determines the focal length of a lens based on its shape and the refractive index of the material. For a lens with surfaces having radii of curvature denoted by \( R_1 \) and \( R_2 \), the equation is written as follows:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:

• \( n \) is the refractive index of the lens material.
• \( R_1 \) is the radius of curvature of the first lens surface.
• \( R_2 \) is the radius of curvature of the second lens surface.

For a planoconvex lens, one surface is flat, essentially making \( R_1 = \infty \).
This simplifies the equation as: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{\infty} - \frac{1}{R_2} \right) \]
This explains why \( R_1 \) does not contribute to the equation when dealing with a flat surface.

Image Formation

Image formation through a lens involves light rays bending (refracting) as they pass through the lens. The way a lens bends light depends on its curvature and material. In optics, the focal length of the lens determines how these rays converge to form an image.

The position of an image formed by a lens is calculated using the lens equation:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
In this formula:

• \( f \) is the focal length.
• \( u \) is the distance from the object to the lens (negative if the object is on the left of the lens).
• \( v \) is the distance from the lens to the image; solving for \( v \) gives the image location.

If \( v \) is positive, the image is real and located on the side of the lens opposite to the object. If \( v \) is negative, the image is virtual and located on the same side as the object.

Magnification

Magnification describes how much larger or smaller the image is compared to the object size. It is calculated as follows:
\[ M = -\frac{v}{u} \]
Where:

• \( M \) is the magnification.
• \( v \) is the image distance.
• \( u \) is the object distance.

The negative sign indicates that if the image is inverted compared to the object. A magnification greater than 1 means the image is larger than the object. In contrast, a value between 0 and 1 indicates a smaller image.

When an image is real, it is inverted due to the negative magnification. For our insect example, a magnification of 2.82 means the image is nearly 3 times larger but upside down compared to the original object.

Planoconvex Lens

A planoconvex lens is characterized by one flat surface and one convex surface. It is used primarily for focusing light, much like a magnifying glass. The convex side causes parallel light rays to converge to a point - the focal point.

In a typical optical setup, planoconvex lenses are used where light needs to be both convergent and parallel when entering or exiting a system.

• The plano side is usually oriented towards the more sensitive part of a system (like a sensor) to minimize spherical aberration.
• The convex side faces incoming light waves.

The shape of the planoconvex lens and its high refractive index makes it especially efficient in developing clear and bright images with minimal distortion.