A compound microscope has a magnifying power of 100 when the image is formed at infinity,.The object has a focal length of 0.5cm and tube length is 6.5 cm .Find the focal length of eye piece

To determine the focal length of the eyepiece in a compound microscope, we can use the formula for magnifying power and some basic optics principles. The magnifying power (M) of a compound microscope is given by the product of the magnification produced by the objective lens (Mo) and the eyepiece lens (Me). In this case, we know that the total magnifying power is 100, the focal length of the objective lens (f_o) is 0.5 cm, and the tube length (L) is 6.5 cm.

Understanding the Components

In a compound microscope, the objective lens is responsible for creating a real image of the object, which is then magnified by the eyepiece. The magnifying power can be expressed as:

• M = Mo × Me

Where:

• Mo = L / f_o (the magnification of the objective lens)
• Me = 1 + (D / f_e) (the magnification of the eyepiece lens)

Here, D is the near point distance of the eye, typically taken as 25 cm, and f_e is the focal length of the eyepiece.

Calculating the Objective Magnification

First, we can calculate the magnification of the objective lens:

• Mo = L / f_o = 6.5 cm / 0.5 cm = 13

Finding the Eyepiece Magnification

Now that we have Mo, we can find Me using the total magnifying power:

• 100 = 13 × Me

Solving for Me gives:

• Me = 100 / 13 ≈ 7.69

Using the Eyepiece Magnification Formula

Next, we can use the formula for the eyepiece magnification:

• Me = 1 + (D / f_e)

Substituting the value of Me we found:

• 7.69 = 1 + (25 cm / f_e)

Rearranging this equation gives:

• 6.69 = 25 cm / f_e

Now, solving for f_e:

• f_e = 25 cm / 6.69 ≈ 3.74 cm

Final Result

The focal length of the eyepiece is approximately 3.74 cm. This value is essential for understanding how the eyepiece contributes to the overall magnification of the microscope, allowing for detailed observation of small objects.