A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
(magnifying_power) 88
(separation) 9.47 cm (approximately 9.5 cm)
Explanation
This problem requires applying the lens formula to both the objective and eyepiece lenses of a compound microscope. First, we find the image distance for the objective lens using the given object distance. Then, we determine where the object must be placed relative to the eyepiece to form an image at the near point (25 cm). The lens separation is the sum of the objective image distance and the eyepiece object distance. Finally, the magnifying power is the product of the magnifications produced by each lens.
Solution Steps
Step 1: Find image distance for objective lens using lens formula. Given: f_o = 8.0 mm = 0.8 cm, u_o = -9.0 mm = -0.9 cm. Using 1/f_o = 1/v_o - 1/u_o, we get 1/0.8 = 1/v_o - 1/(-0.9). Solving: 1/v_o = 1/0.8 - 1/0.9 = (0.9-0.8)/0.72 = 0.1/0.72. Therefore, v_o = 7.2 cm.
Step 2: Find object distance for eyepiece. The final image is at near point D = 25 cm, so v_e = -25 cm (virtual image). Given f_e = 2.5 cm. Using lens formula: 1/f_e = 1/v_e - 1/u_e. Substituting: 1/2.5 = 1/(-25) - 1/u_e. Solving: -1/u_e = 1/2.5 + 1/25 = 10/25 + 1/25 = 11/25. Therefore, u_e = -25/11 = -2.27 cm.
Step 3: Calculate lens separation. The image formed by objective acts as object for eyepiece. Separation = v_o + |u_e| = 7.2 + 2.27 = 9.47 cm.
Step 4: Calculate magnifying power. Magnification by objective: m_o = v_o/|u_o| = 7.2/0.9 = 8. Magnification by eyepiece: m_e = 1 + D/f_e = 1 + 25/2.5 = 11. Total magnifying power: m = m_o × m_e = = 88.