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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?

Short Answer

Expert verified
Set the objective lens-to-image distance at 6.25 cm for 30X magnification.

Step by step solution

01

Understanding the Relationship between Components

A compound microscope consists of an objective lens and an eyepiece (or ocular lens). The total magnification, or angular magnification \( M \), is the product of the magnifications of the objective \( M_o \) and the eyepiece \( M_e \). For a microscope, the magnification is given by \( M = M_o \times M_e \). We are given that \( M = 30X \).
02

Calculating Objective Magnification

The magnification of the objective lens \( M_o \) is defined as \( M_o = \frac{v_o}{f_o} \), where \( v_o \) is the image distance for the objective and \( f_o = 1.25 \text{ cm} \) is its focal length. We need to rearrange this equation to find \( v_o \) since it is not directly given: \( v_o = M_o \times f_o \).
03

Calculating Eyepiece Magnification

The magnification of the eyepiece \( M_e \) is given by \( M_e = 1 + \frac{d}{f_e} \), where \( d \) is the least distance of distinct vision (typically about \( 25 \text{ cm} \) for normal vision) and \( f_e = 5 \text{ cm} \) is the focal length of the eyepiece.
04

Combining Magnifications for Desired Setup

Using \( 30X = M_o \times M_e \), rearrange to get \( M_o \) and \( M_e \). Since \( M_o = \frac{v_o}{f_o} \) and \( M_e = 5 \), we derive \( v_o = M_o \times f_o \). By knowing that typical vision \( d = 25 \text{ cm} \), we solve for \( v_o \) and \( f_o \) to satisfy \( M = 30X \).
05

Calculating Specific Distances and Setting Up

Substitute into \( 30X = \frac{v_o}{1.25} \times (1 + \frac{25}{5}) \) to find \( v_o \). Simplified, this becomes \( 30 = \frac{v_o}{1.25} \times 6 \). Hence, \( v_o = \frac{30 \times 1.25}{6} = 6.25 \text{ cm} \). So, set the objective lens-to-image distance at 6.25 cm.

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Key Concepts

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

Angular Magnification
Angular magnification describes how much larger an object appears when viewed through an optical instrument compared to when seen with the naked eye. In a compound microscope, angular magnification is the product of the magnifications of both the objective lens and the eyepiece. This total magnification helps us see tiny details of microscopic objects. It allows scientists and researchers to study cells and microorganisms closely.
For instance, if a microscope has an angular magnification of \(30X\), it means images are magnified 30 times larger than what the naked eye would perceive. In our exercise, the goal is to achieve this level of magnification, which involves calculating individual parts' contributions to this total increase. Understanding angular magnification is key in effectively setting up and using a compound microscope.
Objective Lens
The objective lens in a compound microscope is responsible for gathering light from the specimen and creating a magnified image within the tube. This lens is placed close to the specimen and is a critical part of determining overall magnification.
The focal length of the objective lens, \(f_o\), directly affects how the light converges to form this first image. In our exercise, the objective lens has a focal length of \(1.25 \text{ cm}\) which helps determine how far the light pathway (or image distance, \(v_o\)) within the microscope should be set to properly magnify the image.
  • To calculate magnification by the objective lens, use \(M_o = \frac{v_o}{f_o}\).
  • This requires knowing the distance \(v_o\), where the image comes to focus.
In this setup, knowing the objective lens specifications helps in adjusting the microscope to achieve the desired total magnification.
Eyepiece
The eyepiece, also known as the ocular lens, is the part of the microscope you look through. It further magnifies the image produced by the objective lens. This additional magnification is what contributes the most to the total angular magnification achieved by the microscope.
The eyepiece in our exercise has a focal length of \(5 \text{ cm}\). The magnification by the eyepiece, \(M_e\), is calculated using the formula \(M_e = 1 + \frac{d}{f_e}\), where \(d\) represents the least distance of distinct vision, typically \(25 \text{ cm}\).
  • The eyepiece's role is crucial because it fine-tunes the image through further enlargement.
  • Even slight adjustments can significantly affect what you see and the ease of your observations.
Thus, understanding how the eyepiece works and adjusting its settings are essential for effectively using a compound microscope.
Focal Length
Focal length is a key concept in understanding how both the objective lens and the eyepiece work in a magnification system like a microscope. It determines how the lenses focus light and form images, which is critical for achieving clear and detailed magnified views.
In a compound microscope setup, the focal length of the objective lens (\(1.25 \text{ cm}\) in this exercise) dictates how closely the lens should be to the specimen for optimal image gathering. Meanwhile, the eyepiece's focal length (\(5 \text{ cm}\)) guides the researcher in setting up the eyepiece to ensure the image is properly enlarged.
  • Objective focal length: Controls the formation of the primary image by focusing light from the specimen.
  • Eyepiece focal length: Helps in adjusting the view for a clear and magnified output.
Knowing these focal lengths allows you to calculate the distances needed for setting up the microscope, ensuring every tiny detail of the specimen is visible and analyzable.

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Most popular questions from this chapter

A small candle, \(2.5 \mathrm{~cm}\) in size is placed at \(27 \mathrm{~cm}\) in front of a concave mirror of radius of curvature \(36 \mathrm{~cm}\). At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

What is the focal length of a convex lens of focal length \(30 \mathrm{~cm}\) in contact with a concave lens of focal length \(20 \mathrm{~cm} ?\) Is the system a converging or a diverging lens? Ignore thickness of the lenses.

{ (a) A giant refracting telescope at an observatory has an objective }\end{array}\( lens of focal length \)15 \mathrm{~m}\(. If an eyepiece of focal length \)1.0 \mathrm{~cm}\( is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is \)3.48 \times 10^{6} \mathrm{~m}\(, and the radius of lunar orbit is \)3.8 \times 10^{8} \mathrm{~m}$.

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.

At what angle should a ray of light be incident on the face of a prism of refracting angle \(60^{\circ}\) so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is \(1.524 .\)

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