The microscope has a converging lens (the eyepiece) with a focal length of 2.50 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 . When you place the sample to be examined at a distance of 1.30 from the objective, at what length will you need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye?
The microscope available in your biology lab has a converging lens (the eyepiece) with a focal length of 2.50 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 . When you place the sample to be examined at a distance of 1.30 from the objective, at what length will you need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye?
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?
To find the length at which you need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye, you can use the lens formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens
- v is the distance of the image from the lens
- u is the distance of the object from the lens
In this case, the focal length of the eyepiece lens (converging lens) is 2.50 and the focal length of the objective lens (converging lens) is 1.00.
The objective lens forms an intermediate image that is then magnified by the eyepiece lens. When viewing the image with a completely relaxed eye, the final image appears at infinity (as if it's at the least distance of distinct vision).
Given:
- The distance of the object from the objective lens, u = 1.30
- The distance of the final image from the eyepiece lens, v = infinity
Substituting these values into the lens formula and solving for v:
1/f_eyepiece = 1/v - 1/u
Since v is infinity, 1/v is 0:
1/f_eyepiece = 0 - 1/1.30
Simplifying:
1/f_eyepiece = -1/1.30
To find the reciprocal of -1/1.30, invert it:
1/f_eyepiece = -1.30
So, the focal length of the eyepiece lens should be -1.30.
To find the length at which you need to adjust the tube of the microscope, you can use the formula:
1/f_system = 1/f_eyepiece + 1/f_objective
Substituting the values:
1/f_system = 1/-1.30 + 1/1.00
Simplifying:
1/f_system = -0.7692 + 1.0000
1/f_system = 0.2308
To find the length, you can use the formula:
magnification = f_system / f_objective
Solving for f_system:
f_system = magnification * f_objective
Since the microscope is in focus with a completely relaxed eye, the magnification is 1:
f_system = 1 * 1.00
f_system = 1.00
Therefore, the focal length of the system is 1.00.
Now, substitute this value back into the lens formula to find the length:
1/f_system = 1/v - 1/u
1/1.00 = 0 - 1/1.30
Simplifying:
1/1.00 = -1/1.30
The reciprocal of -1/1.30 is:
1.300
So, the length at which you need to adjust the tube of the microscope to view the sample in focus with a completely relaxed eye is 1.300 units.