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The microscope available in your biology lab has a converging lens (the eyepiece) with a focal length of 2.50 cm mounted on one end of a tube of adjustable length. At the other end is another converging lens (the objective) that has a focal length of 1.00 cm . When you place the sample to be examined at a distance of 1.30 cm from the objective, at what length l will you need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye

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Answer:

6.83 cm

Explanation:

Given data:

Focal length of the eye piece, [tex]f_e[/tex] = 2.50 cm

Focal length of the converging lens, f = 1.00 cm

distance of the object, p = 1.30 cm

Now,

we have the lens equation as:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

q is the distance of the image

thus,

on substituting the values in the above equation, we get

[tex]\frac{1}{1.00}=\frac{1}{1.30}+\frac{1}{q}[/tex]

or

q = 4.33 cm

now, the image is formed at the focal point of the eye piece,

therefore, the distance between the objective and the eyepiece, d = [tex]f_e[/tex] + q = 2.50 cm + 4.33 cm

or

d = 6.83 cm

onyebuchinnaji

The length you will need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye is 6.83 cm.

Image distance

The position of the image in the lens can be determined by using lens formula as shown below;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/1 - 1/1.3

1/v = 0.231

v = 1/0.231

v = 4.33 cm

Adjusted length of the microscope tube

The length you will need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye is calculated as follows;

adjusted length = 4.33 cm + 2.5 cm = 6.83 cm.

Learn more about lens formula here: https://brainly.com/question/25779311

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