Answer:
The correct option is:
c. The image is virtual and is near the focal point of the objective.
Step-by-step explanation:
In a compound microscope, there are two main lenses: the objective lens and the eyepiece (ocular). The objective lens forms a real and magnified image of the object, which is located near its focal point. This real image then serves as the object for the eyepiece lens, which further magnifies the image and makes it virtual for the observer.
Given the information provided:
Distance between lenses (tube length) = 20 cm
Focal length of the eyepiece (ocular) = 6 mm
Total magnification = 400
The total magnification of a compound microscope is given by the product of the magnification of the objective and the magnification of the eyepiece:
Total Magnification = Magnification (Objective) × Magnification (Eyepiece)
Given that the total magnification is 400, and the eyepiece focal length is 6 mm, the magnification of the objective can be calculated as follows:
Magnification (Objective) = Total Magnification / Magnification (Eyepiece)
Magnification (Objective) = 400 / (25 cm / 6 mm)
Magnification (Objective) = 96
This means the objective lens provides a magnification of 96x. Since the user has a near point at 25 cm, it implies that the final image should be at this distance to be comfortably viewed.
The image formed by the objective lens is real and magnified, but it's not in the same direction as the original object (it's inverted). This real image is located near the focal point of the objective lens. This image then serves as the object for the eyepiece lens, which further magnifies it and makes it virtual for the observer to view comfortably.
So, option c is the correct one: The image is virtual and is near the focal point of the objective.