Final answer:
The magnification produced by a compound microscope with the given powers of the objective (100d) and the eyepiece (10d) when the final image is at the least distance of distinct vision (25 cm) is 105.
Step-by-step explanation:
The question asks for the magnification of a compound microscope when the final image is at the least distance of distinct vision, given the power of the objective and the eyepiece. The magnification to be calculated consists of two parts: the magnification by the objective and the magnification by the eyepiece.
The magnification (m) of a compound microscope is the product of the magnification of the objective lens (mobj) and the magnification of the eyepiece (meye). The power of a lens (P) is the reciprocal of its focal length (f) in meters. That is, P = 1/f.
To find mobj, we use the formula mobj = tube length / focal length of objective.
Given that the power of the objective is 100 diopters (d), its focal length is 1/100 meters, or 0.01 meters (1 cm).
Therefore, mobj = 30 cm / 1 cm = 30.
The magnification of the eyepiece (meye) can be found using the formula meye = 1 + (least distance of distinct vision / focal length of eyepiece). The power of the eyepiece is given as 10d, so its focal length is 1/10 meters, or 0.1 meters (10 cm).
Therefore, meye = 1 + (25 cm / 10 cm)
= 3.5.
To find the total magnification, we multiply the magnifications of the objective and the eyepiece. Thus, the total magnification is m = mobj × meye
= 30 × 3.5
= 105.
In summary, the magnification produced by the microscope when the final image is at the least distance of distinct vision (25 cm) is 105.