In a compound microscope, the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:
Question
In a compound microscope, the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:
Solution
To find the magnifying power of the compound microscope, we can use the formula:
Magnifying Power = (1 + (D/F)) * (1 + (D'/F'))
Where: D = distance of the object from the objective lens F = focal length of the objective lens D' = distance of the final image from the eye piece F' = focal length of the eye piece
Given: Focal length of the objective lens (F) = 1.2 cm Focal length of the eye piece (F') = 3.0 cm Distance of the object from the objective lens (D) = 1.25 cm Distance of the final image from the eye piece (D') = infinity
Since the final image is formed at infinity, we can assume that D' is very large compared to F'. Therefore, we can approximate the formula as:
Magnifying Power = (1 + (D/F))
Substituting the given values:
Magnifying Power = (1 + (1.25/1.2))
Calculating the value:
Magnifying Power = (1 + 1.0417)
Magnifying Power = 2.0417
Therefore, the magnifying power of the compound microscope is approximately 2.0417.
Similar Questions
To make a compound microscope you can add a 2nd lens to the simple magnifier arrangement. The magnification produced by the objective lens (m) is multiplied by the magnification of the simple magnifier (Me), so the total magnifying power (MP) is increased. That is: MP = m Me.Calculate how far away from the objective lens you need to place the object to be viewed to produce a magnifcation of m = - 20, if the focal length of the lens is fo = 10.5 cm. Give answer to 3 significant figures.Answer:
For a compound microscope the magnifying power (MP) is the objective magnification (m) multiplied by the eyepiece magnification (Me). Generally two small focal length lenses are used with around the same value of f. A simple magnifier has been set up with a virtual image produced at 25 cm away from the lens of focal length fe. The object is placed at the required distance 6.3 cm away and can be viewed. Another lens is now added as the objective in a compound microscope. The first lens is the eyepiece. The objective lens has focal length of fo = 9.5 cm and for a magnification of m = -2 it can be calculated that the object now needs to be placed 14.25 cm away from this lens. How far apart must the two lenses be set to construct a compound microscope with the virtual image produced at 25 cm?Give your answer to 4 significant figures.
Find the magnifying power of a compound microscope whose objective and eyepiece are of focal lengths 4.0 cm and 6.0 cm respectively and the object is placed 5.0 cm beyond the objective. Assume that the final image is formed at the least distance of distinct vision (25 cm)
What is the magnification of the low power objective lens of the compound microscope?Group of answer choices40x4x100x10x
A student's microscope has an ocular lens with a 10x magnification. Calculate the total magnification, if the objective lens is set to 30x.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.