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Chapter 24: Problem 94

A compound microscope has magnifying power as 32 and magnifying power of eye- piece is \(4 ;\) then the magnifying power of the objective is (a) 8 (b) 10 (c) 6 (d) 12

Short Answer

Expert verified
The magnifying power of the objective is 8.

Step by step solution

01

Recall the Formula for Total Magnification

The total magnification of a compound microscope, denoted as \( M \), can be calculated by multiplying the magnifications of the objective \( M_o \) and the eyepiece \( M_e \). The formula is given by:\[ M = M_o \times M_e \]
02

Identify Given Values

We know from the problem that the total magnifying power \( M \) is 32 and the magnifying power of the eyepiece \( M_e \) is 4.
03

Solve for Magnification of Objective

Substitute the known values into the formula: \[ 32 = M_o \times 4 \] To solve for \( M_o \), divide both sides of the equation by 4:\[ M_o = \frac{32}{4} = 8 \]

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Key Concepts

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

Total Magnification Formula
Understanding the total magnification of a compound microscope is crucial to mastering its application. The formula used to calculate the total magnification is: \[ M = M_o \times M_e \]where:
  • \( M \) is the total magnification.
  • \( M_o \) is the magnifying power of the objective lens.
  • \( M_e \) is the magnifying power of the eyepiece.
The compound microscope employs two sets of lenses to achieve high levels of magnification. Each lens contributes to making the observed specimen appear much larger than its actual size. This equation highlights how both lenses work together to magnify objects that are incredibly small. Understanding this concept is the cornerstone to solving any problem that involves calculating the magnification provided by a compound microscope.
Objective Magnifying Power
The objective lens is situated closer to the specimen you are observing. It plays a vital role in magnifying the image before it reaches the eyepiece. In the context of the problem, the goal is to determine the objective magnifying power when given the total magnifying power and the eyepiece magnifying power.Given the formula:\[ M_o = \frac{M}{M_e} \]This equation is used to isolate and compute the magnification provided by the objective lens. By substituting the known values into this formula, we are able to solve for the unknown variable. For instance, if the total magnification is 32 and the eyepiece's magnification is 4, then the objective's magnification would be:\[ M_o = \frac{32}{4} = 8 \]This means the objective lens alone increases the apparent size of the specimen by eight times. The objective lens typically provides the primary magnification and is available in several power levels to suit different investigative needs.
Eyepiece Magnifying Power
The eyepiece lens, also known as the ocular lens, is the lens you look through at the top of the microscope. It provides additional magnification on top of what the objective lens has already accomplished. Eyepiece magnifying power is crucial because it amplifies the image formed by the objective lens. It often has lower magnifying power compared to objectives, such as 5x or 10x. In our scenario, the eyepiece has a magnification factor of 4, which implies that it magnifies the image formed by the objective lens fourfold. When coupled with the correct objective magnification, the eyepiece allows you to observe minute details that are not visible to the naked eye or with a single-lens magnifier. This step in the magnification process is vital for viewing fine structural details that could be the difference between identifying a cellular feature or missing it entirely.

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Most popular questions from this chapter

A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length \(10 \mathrm{~cm}\). The diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m}\) and its mean distance from the earth is \(1.5 \times 10^{11} \mathrm{~m}\). What is the diameter of the sun's image on the paper? (a) \(6.5 \times 10^{-5} \mathrm{~m}\) (b) \(12.4 \times 10^{-4} \mathrm{~m}\) (c) \(9.2 \times 10^{-4} \mathrm{~m}\) (d) \(6.5 \times 10^{-4} \mathrm{~m}\)

A convex lens, a glass slab, a glass prism and a spherical solid ball have been prepared from the same optically transparent material. Dispersive power will be possessed only by (a) the glass slab and the prism (b) the lens and the solid ball (c) the prism only (d) all of the above

A beam of light consisting of red, green and blue colours is incident on a right-angled isosoles prism. The refractive indices of the material of the prism for the above red, green and blue wavelengths are \(1.39,1.44\) and \(1.47\) respectively. The prism will (a) separate part of the red colour from the green and blue colours (b) separate part of the blue colour form the red and green colours (c) separate all the three colours from one another (d) not separate even partially any colour from the other two colours

A man suffering from short sight is unable to see objects distinctly at a distance greater than \(2 \mathrm{~m}\). The power of lens required to correct this defect should be (a) \(-0.5 \mathrm{D}\) (b) \(-2 \mathrm{D}\) (c) \(+0.5 \mathrm{D}\) \((\mathrm{d})+2 \mathrm{D}\)

The reflective surface is given by \(y=2 \sin x\). The reflective surface is facing positive axis. What is the least value of coordinates of the point where a ray parallel to positive \(x\) -axis becomes parallel to positive \(y\) -axis after reflection? (a) \(\left(\frac{\pi}{3}, \sqrt{3}\right)\) (b) \(\left(\frac{\pi}{2}, \sqrt{2}\right)\) (c) \(\left(\frac{\pi}{3}, \sqrt{2}\right)\) (d) \(\left(\frac{\pi}{4}, \sqrt{3}\right)\)
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