Since it still provides a good amount of magnification at a good distance from the slide, there is a limited risk of it breaking the glass and potentially ruining the sample. That’s why this objective lens is often preferred before going for a high powered lens.

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

The majority of compound microscopes come with interchangeable objective lenses, which have different magnification powers. This commonly includes 4x, 10x, 40x, and 100x objective lenses.

Objective lens function in microscopeand itsfunction

As previously mentioned, the ocular or eyepiece lens is located at the top of the eyepiece tube and is where you position your eye to observe the specimen. The ocular lens typically has a low magnification (10x) and works in combination with the objective lens to achieve a greater magnification power.

This is referred to as the high powered objective lens since it is ideal for observing the small details within a specimen sample. The total magnification for this lens is equal to 400x magnification (10x eyepiece lens x the 40x objective equals 400).

The objective lens and the ocular or eyepiece lens are in combination responsible for magnification of the specimen being observed.

Resolving power is also a very important metric since magnification power is of little importance if the resolution is not high. Resolution is defined as the ability to distinguish 2 points as two points.

Often overlooked is the cleanliness of your optics. Daily use in any environment will attract dust and small debris, and when handling your lens, oils from your body can be transferred. This is particularly the case around the eyepiece.

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

The lenses of the microscope are fundamental to its function as they provide the magnification power that allows the microscopic specimen to be seen or observed in greater detail. The two main types of lenses found in light microscopes today are called the objective lens and the ocular lens, which is also called the eyepiece.

5. A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

The majority of light microscopes have an objective lens of some kind, including both compound microscopes and stereo microscopes. Both of these types of microscopes also have an eyepiece or ocular lens.

9. A 2.8-cm diameter coin is placed a distance of 25.0 cm from a double concave lens that has a focal length of -12.0 cm. Determine the image distance and the diameter of the image.

Ocularlens microscope function

As a demonstration of the effectiveness of the lens equation and magnification equation, consider the following sample problem and its solution.

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 35.5 cm from a diverging lens having a focal length of 12.2 cm, then the image will be upright, 1.02-cm tall and located 9.08 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. Diverging lenses always produce images that are upright, virtual, reduced in size, and located on the object's side of the lens.

To determine the image distance, the lens equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

This objective lens is the next lowest powered and is often the most helpful when it comes to analyzing glass slide samples. The total magnification for this lens is equal to 100x magnification (10x eyepiece lens x the 10x objective equals 100).

2. Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double convex lens having a focal length of 15.0 cm.

Because glass and air have different refractive indexes, light bends at different angles when it passes through each of them. When using the 4x, 10x, 40x objective lenses, the light refraction that occurs when looking through the lens to the specimen on the glass slide is not very noticeable. However, when using the higher power objective lenses, for example the 100x, the light refraction is much more obvious.

High powerobjective microscope function

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

From the calculations in the second sample problem it can be concluded that if a 4.00-cm tall object is placed 8.30 cm from a double convex lens having a focal length of 15.2 cm, then the image will be enlarged, upright, 8.81-cm tall and located 18.3 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in front of the focal point (i.e., the object distance is less than the focal length) and the image is located behind the lens. This falls into the category of Case 5: The object is located in front of F (for a converging lens).

3. Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double convex lens having a focal length of 15.0 cm.

7. A double concave lens has a focal length of -10.8 cm. An object is placed 32.7 cm from the lens's surface. Determine the image distance.

Stagemicroscope function

For example, if you are looking down a microscope, the resolution power relates to the space you can see between two points. A very low resolution would result in a blurred image and would prevent proper observation of the specimen.

The objective lens is at the bottom of the eyepiece tube and is responsible for both total magnification of the specimen, as well as the resolving power of the microscope.

6. ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double convex lens. Determine the image distance and the focal length of the lens.

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

Light microscopes are relatively complex pieces of equipment in nature with multiple different parts, some which are more complex than others.

1. Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double convex lens having a focal length of 15.0 cm.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

Like many mathematical problems in physics, the skill is only acquired through much personal practice. Perhaps you would like to take some time to try the following problems.

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

The magnification of an image is both the hi / ho ratio and the -di / do ratio. Setting the -di / do ratio equal to -2 allows one to determine the image distance:

8. Determine the focal length of a double concave lens that produces an image that is 16.0 cm behind the lens when the object is 28.5 cm from the lens.

This objective is often referred to as the scanning objective lens since the low power provides enough magnification to give the observer a good overview of the entire slide and sample.

10. The focal point is located 20.0 cm from a double concave lens. An object is placed 12 cm from the lens. Determine the image distance.

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:

Combined with the eyepiece lens, this lens will provide the lowest magnification power. For example, 10x eyepiece lens, multiplied by the 4x objective lens gives a total magnification of 40x.

What isobjective lens in microscope

A drop of special oil which has a similar refractive index to glass, is placed on the cover slip over the specimen.  The oil immersion objective lens is immersed in the oil, rather than air, enabling a clear image of the specimen.

Functionof condenserin microscope

While the total magnification is determined by both the objective and ocular lens, the resolution is determined by the objective lens alone.

4. Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double convex lens having a focal length of 15.0 cm.

The negative value for image distance indicates that the image is a virtual image located on the object's side of the lens. Again, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always means the image is located on the object's side of the lens. Note also that the image height is a positive value, meaning an upright image. Any image that is upright and located on the object's side of the lens is considered to be a virtual image.

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

The negative values for image distance indicate that the image is located on the object's side of the lens. As mentioned, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always indicates the existence of a virtual image located on the object's side of the lens. In the case of the image height, a positive value indicates an upright image.

Objective lens function in microscopeand their functions

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

Ensuring your lenses are are kept clean will increase the performance and clarity of your microscope’s images.  There are many products on the market but nothing specifically made for microscopes.  We found a good quality Digital Camera Cleaning Kit was a great option, something with wipes for removing grease and oils and a puffer bottle for blowing away dust.

Compoundobjective lens function in microscope

The ocular lens is positioned at the top of the optical tube, while the objective lens is positioned at the bottom. Both of these lenses have important roles in magnification, but the objective lens also has other defined roles, such as resolving power.

This objective lens will achieve the greatest magnification and has a total magnification of 1000x (10x eyepiece lens x the 100x objective equals 1000).

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm from a double convex lens having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall and located 22.8 cm from the lens. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located beyond the 2F point (which would be two focal lengths from the lens) and the image is located between the 2F point and the focal point. This falls into the category of Case 1: The object is located beyond 2F for a converging lens.

The negative values for image height indicate that the image is an inverted image. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image height, a negative value always indicates an inverted image.