Final answer:
The magnification produced by a 0.150 cm-focal length microscope objective that is 0.155 cm from the object is approximately -3.00, meaning it is inverted. When combined with an 8x eyepiece, the overall magnification is -24, also inverted.
Step-by-step explanation:
To find the magnification produced by a 0.150 cm-focal length microscope objective that is 0.155 cm from the object being viewed, we use the lens formula: 1/f = 1/d_o + 1/d_i, where f stands for the focal length, d_o is the object distance (given), and d_i is the image distance. However, for the purpose of this question, we can use the approximation that magnification (M) is equal to the image distance over the object distance because the object is very close to the focal point. In this case, since the object distance (0.155 cm) is nearly the focal length of the objective (0.150 cm), the magnification can be approximated as −f/(d_o − f), considering the negative sign because the image is inverted.
The magnification produced by the microscope objective will therefore be M ≈ -0.150/(0.155-0.150) = -3.00. Please note, the negative sign denotes that the image is inverted.
For part (b), the overall magnification when an 8x eyepiece is used is the product of the magnifications of the objective lens and the eyepiece. Since the objective provides a magnification of -3.00 and the eyepiece an additional magnification of 8.00, the overall magnification is -3.00 x 8.00 = -24. This implies that the image will appear 24 times larger than the actual object and will be inverted.