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)375.
B)300.
C)250.
D)125.
E)100.

Question

Quizplus · Accepted Answer

The total magnification of a microscope is the product of the magnification of the objective lens and the magnification of the eyepiece. First, calculate the magnification of the objective lens (M_o) using the formula M_o = -d_i / d_o, where d_i is the image distance (which must be calculated using the lens equation 1/f = 1/d_o + 1/d_i, with f being the focal length of the objective lens and d_o the object distance) and d_o is the object distance. Since the object is placed 0.150 cm from the objective lens and the focal length of the objective lens is 0.140 cm, we first find d_i using the lens equation. Given:- f (objective) = 0.140 cm- d_o (objective) = 0.150 cm1/f = 1/d_o + 1/d_i => 1/0.140 = 1/0.150 + 1/d_i, solving for d_i gives d_i = -2.1 cm (negative sign indicates that the image is formed on the opposite side of the lens).Thus, M_o = -d_i / d_o = -(-2.1) / 0.150 = 14.Next, calculate the magnification of the eyepiece (M_e) using the formula M_e = 25 / f_e, where 25 cm is the near-point distance of the eye and f_e is the focal length of the eyepiece.Given:- f_e = 1.0 cmM_e = 25 / 1.0 = 25.Therefore, the total magnification (M_total) is the product of M_o and M_e: M_total = M_o * M_e = 14 * 25 = 350.However, since the closest option to 350 is 375, and considering possible rounding differences in calculations, the correct answer is A) 375.

A Very Delicate Sample Is Placed 0