A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm, and that of the eyepiece is 1.0 cm. The near-point distance of the
Person using the microscope is 25.0 cm. The magnitude of the final magnification of the
Microscope is closes to
A) 125.
B) 375.
C) 250.
D) 300.
E) 100.

Question

Quizplus · Accepted Answer

The total magnification of a microscope is the product of the magnification of the objective lens ($M_{obj}$) and the magnification of the eyepiece ($M_{eye}$). The magnification of the objective lens can be calculated using the formula $M_{obj} = \frac{-d_i}{d_o}$, where $d_i$ is the image distance (for the objective lens) and $d_o$ is the object distance. Since the object is placed just outside the focal point of the objective lens, the image produced by the objective lens will be on the opposite side and magnified. The image distance ($d_i$) can be found using the lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, where $f$ is the focal length of the lens.Given that the focal length of the objective lens ($f_{obj}$) is 0.140 cm and the object distance ($d_o$) is 0.150 cm, we can find the image distance ($d_i$) for the objective lens. Rearranging the lens equation gives us $d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_o}}$.Substituting the given values, we find $d_i$ for the objective lens. This image acts as a virtual object for the eyepiece. The magnification of the eyepiece ($M_{eye}$) is given by $\frac{25}{f_{eye}}$, where 25 cm is the near-point distance (considered as the image distance for the eyepiece), and $f_{eye}$ is the focal length of the eyepiece.Calculating these values:1. For the objective lens, $f_{obj} = 0.140$ cm, $d_o = 0.150$ cm. Using the lens equation, we find $d_i$ for the objective lens.2. The magnification of the objective lens ($M_{obj}$) is then calculated.3. For the eyepiece, $f_{eye} = 1.0$ cm, and using the given near-point distance, we calculate $M_{eye}$.4. The total magnification is the product of $M_{obj}$ and $M_{eye}$.Given the choices, the correct calculation leads to a total magnification closest to 375, which means the correct answer is B. This involves detailed calculations, but the key point is understanding how to use the lens equation and magnification formulas for both the objective and the eyepiece to find the total magnification of the microscope.

A Very Delicate Sample Is Placed 0