Problem 61 - University Physics with Modern Physics

Question

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{mm}, 4 \mathrm{mm},\) and 1.9 \(\mathrm{mm}\) and with eye-pieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image 120 \(\mathrm{mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

Step-by-Step Solution

Answer

Largest magnification: 621; Smallest magnification: 32.5.

Step 1: Calculate Magnification for Each Objective

The magnification (\(m_o\)) of an objective is given by the formula \(m_o = \frac{l}{f_o} - 1\), where \(l\) is the image distance (120 mm) and \(f_o\) is the focal length of the objective.For \(f_o = 16\) mm: \\(m_o = \frac{120}{16} - 1 = 6.5\).For \(f_o = 4\) mm: \\(m_o = \frac{120}{4} - 1 = 29\).For \(f_o = 1.9\) mm: \\(m_o = \frac{120}{1.9} - 1 \approx 62.1\).

Step 2: Calculate Overall Angular Magnification

The overall angular magnification (\(M\)) is the product of the objective magnification (\(m_o\)) and the eyepiece magnification (\(m_e\)). We need to calculate it for all combinations of objectives and eyepieces.For \(5\times\) eyepiece:- With 16 mm objective: \(M = 6.5 \times 5 = 32.5\)- With 4 mm objective: \(M = 29 \times 5 = 145\)- With 1.9 mm objective: \(M \approx 62.1 \times 5 = 310.5\)For \(10\times\) eyepiece:- With 16 mm objective: \(M = 6.5 \times 10 = 65\)- With 4 mm objective: \(M = 29 \times 10 = 290\)- With 1.9 mm objective: \(M \approx 62.1 \times 10 = 621\)

Step 3: Determine Largest and Smallest Magnification

Compare all results from Step 2:- Largest Magnification: \(621\) (using 1.9 mm objective and 10\(\times\) eyepiece)- Smallest Magnification: \(32.5\) (using 16 mm objective and 5\(\times\) eyepiece)

Key Concepts

Focal Length

The focal length of a lens is crucial to understanding how a microscope magnifies an object. It represents the distance between the lens and the point where parallel rays of light converge to a focus. Short focal lengths mean stronger magnifying power because they converge light more quickly. For example, in the given exercise, the lenses have focal lengths of 16 mm, 4 mm, and 1.9 mm. A shorter focal length like 1.9 mm provides higher magnification because the light focuses more rapidly, making objects appear larger. This is similar to how a zoom lens on a camera works.
To calculate how much an objective lens magnifies, we use the formula: \( m_o = \frac{l}{f_o} - 1 \) where \(l\) is the image distance and \(f_o\) is the focal length. A smaller \(f_o\) results in a higher value for \(m_o\), indicating greater magnification capacity.

Angular Magnification

Angular magnification refers to how much larger an object appears when viewed through a microscope compared to the naked eye. It is determined by multiplying the magnifications of both the objective lens and the eyepiece.
The exercise provided different eyepieces with angular magnifications of \(5 \times\) and \(10 \times\). This means that the eyepiece alone makes an object look 5 to 10 times bigger than it appears without magnification. For instance, if an eyepiece has an angular magnification of \(10 \times\) and it combines with an objective lens that magnifies \(62.1\), the total angular magnification becomes \(621\).
Angular magnification is a product of the optics involved and shows how significantly an image can be expanded to appear larger through a microscope.

Optical Objective

The optical objectives of a microscope are key components that determine how much detail can be observed in a specimen. An objective lens captures light from the sample and creates a real image of the specimen at a certain magnification.
In the case of the exercise, various lenses with focal lengths of 16 mm, 4 mm, and 1.9 mm represent distinct optical objectives. Each creates a different image size before the eyepiece further magnifies it. The shorter focal length objectives provide higher magnification and better detail, essential for examining tiny structures. For instance, the smallest objective (with 1.9 mm focal length) provides the highest magnification due to its smaller focal distance, allowing it to form a larger initial image.
The choice of objective is crucial in microscopy as it directly affects resolution and brightness of the image.