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The magnification of a compound microscope is 30 and the focal length of its eye piece is \(5 \mathrm{~cm}\). The magnification produced by the objective, when the image is to be formed at least distance of distinct vision \((25 \mathrm{~cm})\), is: (a) 5 (b) 6 (c) 8 (d) 10

Short Answer

Expert verified
The magnification produced by the objective is 5, so the answer is (a) 5.

Step by step solution

01

Understanding the Given Information

We are given the total magnification of the compound microscope (\( M = 30 \)) and the focal length of the eyepiece (\( f_e = 5 \text{ cm} \)). We need to find the magnification produced by the objective (\( M_o \)) when the image is formed at the least distance of distinct vision (\( D = 25 \text{ cm} \)).
02

Writing the Formula for Total Magnification

The formula for the total magnification (\( M \)) of a compound microscope is given by:\[M = M_o \times M_e\]where \( M_o \) is the magnification of the objective and \( M_e \) is the magnification of the eyepiece.
03

Calculating Magnification of Eyepiece

The magnification of the eyepiece (\( M_e \)) is calculated using the formula:\[M_e = 1 + \frac{D}{f_e}\]Substitute \( D = 25 \text{ cm} \) and \( f_e = 5 \text{ cm} \) into the formula:\[M_e = 1 + \frac{25}{5} = 1 + 5 = 6\]
04

Solving for the Magnification of the Objective

Using the formula from Step 2 and the value found in Step 3:\[M = M_o \times M_e\]Substitute the known values:\[30 = M_o \times 6\]Rearranging gives:\[M_o = \frac{30}{6} = 5\]
05

Concluding the Solution

The magnification produced by the objective lens is \( M_o = 5 \). Therefore, the correct answer is option (a) 5.

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Key Concepts

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

Magnification
In the world of optics, magnification quantifies how much larger an object appears when viewed through an optical instrument, such as a compound microscope. A compound microscope uses two lenses to magnify the image of a small specimen. The total magnification is the product of the magnification of its individual lenses: the objective lens and the eyepiece. In our problem, the total magnification of the microscope is 30. This indicates how many times larger the image appears compared to the actual object. Mathematically, we express total magnification as:
  • \[ M = M_o \times M_e \]
where \( M \) is the total magnification, \( M_o \) is the magnification of the objective lens, and \( M_e \) is the magnification of the eyepiece.
The greater the magnification, the more detail we can see in the tiny structures of the specimen, allowing for a deeper understanding of its physical properties.
Least Distance of Distinct Vision
The least distance of distinct vision, often denoted as \( D \), is the minimum distance at which the human eye can focus on an object without strain. Generally, this distance is around 25 cm for a person with normal vision.
In the context of microscopy, knowing the least distance of distinct vision helps determine where to position the image formed by a microscope for comfortable viewing. This distance is crucial when calculating the eyepiece magnification, using the formula:
  • \[ M_e = 1 + \frac{D}{f_e} \]
where \( f_e \) is the focal length of the eyepiece. It ensures that the observer can view the magnified image clearly without eye strain.
Focal Length
Focal length is a term used in optics to describe the distance from the lens's center to its focal point, where parallel rays of light converge. It is crucial in determining how much an object is magnified when viewed through a lens.
For a microscope, each lens—objective and eyepiece—has its own focal length, affecting the total magnification of the instrument. The shorter the focal length, the greater the potential magnification because the lens bends light more sharply.
In our exercise, the eyepiece has a focal length of 5 cm. This parameter is used in calculating the eyepiece magnification (\( M_e \)), influencing the overall magnification when combined with the objective lens.
Objective Lens
The objective lens is one of the primary lenses in a compound microscope. Positioned close to the specimen, it is responsible for the initial magnification and focusing of the image.
Different objective lenses offer varying levels of magnification, and they work in tandem with the eyepiece to achieve the total desired magnification. In many microscopes, objective lenses can be switched to adjust the level of detail observed.
  • Low power lenses for general overviews.
  • High power lenses for detailed analysis.
For our specific problem, the task was to find the magnification by the objective lens when the image is viewed at the least distance of distinct vision. Using the total magnification and the magnification of the eyepiece, we determine the objective lens's role in the microscope, promoting a detailed and clear view of minute specimens.

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Most popular questions from this chapter

A compound microscope has an eye piece of focal length \(10 \mathrm{~cm}\) and an objective of focal length \(4 \mathrm{~cm}\). The magnification, if an object is kept at a distance of \(5 \mathrm{~cm}\) from the objective and final image is formed at the least distance of distinct vision \((20 \mathrm{~cm})\), is : (a) 10 (b) 11 (c) 12 (d) 13

The change in the focal length of the lens, if a convex lens of focal length \(20 \mathrm{~cm}\) and refractive index \(1.5\), is immersed in water having refractive index \(1.33\), is : (a) \(62.2 \mathrm{~cm}\) (b) \(5.82 \mathrm{~cm}\) (c) \(58.2 \mathrm{~cm}\) (d) \(6.22 \mathrm{~cm}\)

A concave mirror with its optic axis vertical and mirror facing upward is placed at the bottom of the water tank. The radius of curvature of mirror is \(40 \mathrm{~cm}\) and refractive index for water \(\mu=4 / 3\). The tank is \(20 \mathrm{~cm}\) deep and if a bird is flying over the tank at a height \(60 \mathrm{~cm}\) above the surface of water, the position of image of a bird is: (a) \(3.75 \mathrm{~cm}\) (b) \(4.23 \mathrm{~cm}\) (c) \(5.2 \mathrm{~cm}\) (d) \(3.2 \mathrm{~cm}\)

A light ray strikes a flat glass plate, at a small angle ' \(\theta^{\prime}\). The glass plate has thickness ' \(t\) ' and refractive index ' \(\mu^{\prime}\). What is the lateral displacement ' \(d^{\prime}\) ? (a) \(\frac{t \theta(\mu+1)}{\mu}\) (b) \(\frac{t \theta(\mu-1)}{\mu}\) (c) \(\frac{t}{\theta \mu}(\mu-1)\) (d) \(\frac{\mu}{t \theta}(\mu+1)\)

The human eye can be regarded as a single spherical refractive surface of curvature of cornea \(7.8 \mathrm{~mm}\). If a parallel beam of light comes to focus at \(3.075 \mathrm{~cm}\) behind the refractive surface, the refractive index of the eye is: (a) \(1.34\) (b) 1 (c) \(1.5\) (d) \(1.33\)

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