Problem 78 - Physics for Scientists and Engineers with Modern Physics

Question

(II) The eyepiece of a compound microscope has a focal length of \(2.80 \mathrm{~cm}\) and the objective lens has \(f=0.740 \mathrm{~cm} .\) If an object is placed \(0.790 \mathrm{~cm}\) from the objective lens, calculate (a) the distance between the lenses when the microscope is adjusted for a relaxed eye, and ( \(b\) ) the total magnification.

Step-by-Step Solution

Answer

(a) Distance between lenses: sum of \( d_{io} \) and \( f_e \). (b) Total magnification: \( M_o \times M_e \).

Step 1: Understand the Problem

We are given a compound microscope with details of the eyepiece and objective lens. We need to find the lens separation for a relaxed eye and the total magnification.

Step 2: Calculate the Image Distance from the Objective Lens

For the objective lens, which forms an image of the object, we use the lens formula: \[ \frac{1}{f_o} = \frac{1}{d_o} + \frac{1}{d_{io}} \] where \( f_o = 0.740 \) cm, and \( d_o = 0.790 \) cm. Solving for \( d_{io} \) (the image distance from the objective lens):\[ \frac{1}{0.740} = \frac{1}{0.790} + \frac{1}{d_{io}} \]\[ \frac{1}{d_{io}} = \frac{1}{0.740} - \frac{1}{0.790} \]Calculate \( d_{io} \).

Step 3: Determine the Image Distance from the Eyepiece

For a relaxed eye, the final image is at infinity. Thus, the image formed by the eyepiece is at its focal point, \( f_e = 2.80 \) cm. Hence, the distance of the image formed by the objective lens is the sum of \( d_{io} \) and the focal length of the eyepiece \( f_e \).

Step 4: Calculate the Distance Between the Lenses

Using the results from Steps 2 and 3, we calculate the distance between the two lenses: \( \text{Distance between lenses} = d_{io} + f_e \).

Step 5: Calculate the Magnification of the Objective Lens

The magnification of the objective lens \( M_o \) is given by: \[ M_o = \frac{d_{io}}{d_o} \]. Use \( d_{io} \) obtained in Step 2 and \( d_o = 0.790 \) cm.

Step 6: Calculate the Magnification of the Eyepiece

The magnification of the eyepiece (assuming the final image is at infinity for a relaxed eye) is \( M_e = \frac{25}{f_e} \) where \( f_e = 2.80 \) cm and 25 cm is the assumed near point distance conventionally used.

Step 7: Determine the Total Magnification

The total magnification \( M_t \) of the microscope is the product of the magnifications of the objective and the eyepiece: \[ M_t = M_o \times M_e \]. Substitute the values found in Steps 5 and 6 to find \( M_t \).

Key Concepts

Focal Length

The focal length of a lens is the distance from the lens where a beam of light meets after passing through. It's a crucial factor when determining how a lens will focus light rays. In the case of a compound microscope, different lenses contribute to the overall image creation process.

The objective lens and eyepiece both have specific focal lengths. For example, in the given exercise, the focal length of the eyepiece is given as 2.80 cm, while the objective lens has a much shorter focal length of 0.740 cm. These values influence how the light is bent, focused, and thus how the final image is formed and magnified.

Magnification

Magnification is one of the primary advantages of using a microscope, as it enables us to see tiny details of small objects. It is the process of enlarging the apparent size, not the physical size, of something.

In a compound microscope, magnification occurs in two main stages: through the objective lens and through the eyepiece. The objective lens creates an enlarged image of the object, and the eyepiece further magnifies this image. The total magnification is the product of these two individual magnifications, and in practical terms, it can allow you to see objects at a much larger scale, revealing details not visible to the naked eye.

Lens Formula

The lens formula is critical for determining the relationship between the object distance, image distance, and focal length of a lens. It is given by the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

In this formula, \(f\) represents the focal length, \(d_o\) is the object distance from the lens, and \(d_i\) is the image distance from the lens.

By applying this formula, we can solve for one of these variables if the other two are known, which helps in determining how and where the image will form when using multiple lenses in a compound microscope.

Objective Lens

The objective lens is one of the two main lenses in a compound microscope and is crucial in initial image formation. It is located close to the specimen and is responsible for creating a magnified image of the specimen.

The optical power of the objective lens is characterized by its short focal length, as seen in the exercise where it is 0.740 cm. This short focal length allows it to produce a large amount of initial magnification, setting the stage for further magnification by the eyepiece.

Eyepiece

The eyepiece, also known as the ocular lens, is where you place your eye to observe the image. It further magnifies the image formed by the objective lens, resulting in the final magnified view that you see.

Its magnification power is determined largely by its focal length. The shorter the focal length, the higher the magnification. For example, with a focal length of 2.80 cm as in the exercise, a typical eyepiece can add significant magnification to the initial image created by the objective lens. Moreover, when the microscope is adjusted for a relaxed eye, the eyepiece focuses the image at infinity, allowing for comfortable viewing over longer periods.