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Chapter 9: Problem 11

A compound microscope consists of an objective lens of focal length \(2.0 \mathrm{~cm}\) and an eyepiece of focal length \(6.25 \mathrm{~cm}\) separated by a distance of \(15 \mathrm{~cm}\). How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision \((25 \mathrm{~cm})\), and (b) at infinity? What is the magnifying power of the microscope in each case?

Short Answer

Expert verified
(a) Place the object 1.25 cm from the objective; magnifying power is 39. (b) Place the object 1.75 cm from the objective; magnifying power is 207.375.

Step by step solution

01

Understand the Objective Lens

The object distance for the objective lens is denoted by \( u_o \), and the image distance from the objective lens is \( v_o \). For the objective lens, we use the lens formula: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]where \( f_o = 2.0 \text{ cm} \) is the focal length of the objective lens.
02

Image Formation by the Objective Lens

For the objective, we need the final image either at the least distance of distinct vision or at infinity. Denote the tube length between lenses as \( L \), so the image distance for the eyepiece should be \( 25 \text{ cm} \) for the least distinct vision.Using \( L = v_o + u_e \), where \( u_e \) is the image distance for the eyepiece and \( L = 15 \text{ cm} \), find \( v_o \).
03

Find Object Distance for Objective (Final Image at least distance)

Using the relationship with the eyepiece for least distance: \[ v_o = L - u_e = 15 - 25 = -10 \text{ cm} \]Now, use the lens equation for the objective:\[ \frac{1}{2} = \frac{1}{-10} - \frac{1}{u_o} \]Solve for \( u_o \).
04

Find Object Distance for Objective (Final Image at Infinity)

For the image at infinity, the first image from the objective should be at the focal point of the eyepiece.\( v_o = 15 - 6.25 \text{ cm} = 8.75 \text{ cm} \)Using lens formula for objective:\[ \frac{1}{2} = \frac{1}{8.75} - \frac{1}{u_o} \] and solving for \( u_o \).
05

Calculate Magnifying Power for Both Cases

For least distance: The magnifying power is given by\[ M = \left(1 + \frac{D}{f_e} \right) \left( \frac{v_o}{u_o} \right) \]And for the final image at infinity:\[ M = \frac{v_o}{u_o} \left( \frac{D}{f_e} \right) \] where \( D = 25 \text{ cm} \).
06

Solve for Values

Solve equations from Steps 3 and 4 to find \( u_o \) values for each case.Complete the magnifying power equations using these \( u_o \) and previously found \( v_o \) values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Objective Lens
When using a compound microscope, the objective lens plays a pivotal role in magnification. The objective lens is the one closest to the specimen being observed. It gathers light from the specimen and creates an enlarged image inside the microscope. This lens typically has a small focal length, allowing it to produce a highly magnified image. In our exercise, the focal length of the objective lens is given as \(2.0 \, ext{cm}\). The lens formula, \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \), helps us calculate the image distance \(v_o\) or the object distance \(u_o\), where \(f_o\) is the focal length.
Eyepiece
The eyepiece, unlike the objective lens, is located close to the observer's eye. It acts like a magnifying glass, further enlarging the image produced by the objective lens. In essence, the eyepiece allows you to see this image at a comfortable viewing size. For the compound microscope in the original exercise, the eyepiece's focal length is \(6.25 \, ext{cm}\). The role of the eyepiece is crucial in achieving the total magnification, as it also affects the distance needed between it and the objective lens, known as the tube length.
Magnifying Power
Magnifying power refers to how much larger the microscope makes the specimen appear. This is a key measurement in a compound microscope's efficacy. The full magnification power is calculated by multiplying the magnification of the objective lens by that of the eyepiece. For a compound microscope, we use specific formulas. If the final image is at the least distance of distinct vision, the magnification equation is:
  • \( M = \left(1 + \frac{D}{f_e} \right) \left( \frac{v_o}{u_o} \right) \)
Where \(D\) is the least distance of distinct vision (\(25 \, ext{cm}\)). If the final image is at infinity, the formula simplifies to:
  • \( M = \frac{v_o}{u_o} \left( \frac{D}{f_e} \right) \)
These formulas incorporate both the positioning of the lenses and human vision principles.
Least Distance of Distinct Vision
The least distance of distinct vision is the closest point at which the eye can see an object clearly without strain. This distance can vary slightly between individuals but is typically around \(25 \, ext{cm}\). When using a microscope, achieving an image at this particular distance ensures it's comfortable for viewing through the eyepiece. In the steps given, the \(v_o\) (image distance for the eyepiece) should be set to \(25 \, ext{cm}\) to meet this viewing requirement. This impacts other factors like object placement and overall magnification.
Lens Formula
The lens formula is fundamental in optical calculations. It relates the focal length (\(f\)) of a lens to the object distance (\(u\)) and the image distance (\(v\)). The formula is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]This relationship helps us determine where an object or image should be positioned relative to a lens to achieve clear magnification. In the original problem, different scenarios are considered鈥攔esulting in different positions for the object to achieve an image at a specific distance, either at the least distance of distinct vision or at infinity, using the lens formula as the guiding mathematical principle.

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Most popular questions from this chapter

An angular magnification (magnifying power) of \(30 \mathrm{X}\) is desired using an objective of focal length \(1.25 \mathrm{~cm}\) and an eyepiece of focal length \(5 \mathrm{~cm}\). How will you set up the compound microscope?

For a normal eye, the far point is at infinity and the near point of distinct vision is about \(25 \mathrm{~cm}\) in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

An object of size \(3.0 \mathrm{~cm}\) is placed \(14 \mathrm{~cm}\) in front of a concave lens of focal length \(21 \mathrm{~cm} .\) Describe the image produced by the lens. What happens if the object is moved further away from the lens?

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be \(40^{\circ}\). What is the refractive index of the material of the prism? The refracting angle of the prism is \(60^{\circ} .\) If the prism is placed in water (refractive index \(1.33)\), predict the new angle of minimum deviation of a parallel beam of light.

Use the mirror equation to deduce that: (a) an object placed between \(f\) and \(2 f\) of a concave mirror produces a real image beyond \(2 f\). (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
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