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The focal length of a lens is a fundamental characteristic that determines how strongly the lens converges or diverges light. For a compound microscope, which uses two lenses, knowing the focal length of each lens is crucial to understand how the microscope forms an image.
For the given exercise, the focal length of the objective lens is 8.0 mm, while the focal length of the eyepiece is 2.5 cm. The focal lengths tell us how far the focus of the lens is from the lens itself. The objective lens has a shorter focal length compared to the eyepiece, making it more potent in terms of magnifying power in the microscope.

• Shorter focal length (e.g., 8.0 mm for objective) — Higher magnification.
• Longer focal length (e.g., 2.5 cm for eyepiece) — Lower magnification but wider field of view.

Understanding focal lengths helps us predict how the microscope will perform and affect the image formations ahead.

The lens formula is a mathematical relationship that connects the focal length \((f)\), the object distance \( (u) \), and the image distance \( (v) \). It's essential for determining where the image of an object will form when viewed through a lens. The formula is expressed as:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]In a compound microscope, this formula is exceedingly useful. In the given problem, it was used to find the image distance \( v_o \) of the objective lens. Knowing \( f_o = 8.0 \text{ mm} \) and \( u_o = -9.0 \text{ mm} \), we apply the formula:\[\frac{1}{v_o} = \frac{1}{0.8} - \frac{1}{0.9}\]Solving, you get \( v_o \approx 7.2 \text{ cm} \). By using the lens formula systematically, we can locate the image position necessary for subsequent calculations of lens separation and magnifying power.

Magnifying power in microscopes is a crucial measurement that expresses how much larger an object appears when viewed through the microscope. It combines the powers of the objective lens and the eyepiece. The magnifying power \( M \) is determined using:\[M = \left( \frac{D}{f_e} \right) \left( \frac{v_o}{u_o} \right)\]where \( D \) is the near point distance, usually 25 cm for a normal eye. From the exercise, substitution gives us:\[M = \left( \frac{25}{2.5} \right) \left( \frac{7.2}{0.9} \right) = 10 \times 8 = 80\]This value means that the microscope makes the object appear 80 times larger than with the naked eye. The formula breaks down into two parts:

• \( \frac{D}{f_e} \): Magnification due to the eyepiece.
• \( \frac{v_o}{u_o} \): Magnification due to the objective lens.

Understanding these helps us use lenses effectively to reach specific magnification levels.

Lens separation is about the space between the two lenses in a compound microscope. The separation distance helps determine how the lenses work together to form a clear, magnified image. It’s calculated by simply adding the image distance from the objective to the focal length of the eyepiece.In this exercise, \( v_o = 7.2 \text{ cm} \) and \( f_e = 2.5 \text{ cm} \), so:\[d = v_o + f_e = 7.2 \text{ cm} + 2.5 \text{ cm} = 9.7 \text{ cm}\]This separation ensures that the final image is formed at the observer's near point, helping provide a relaxed viewing experience. Proper lens separation is critical for the effective function of the microscope and ensures the image is correctly focused for user observation.