COMEDK20269 May 2026Morning ShiftPhysicsRay OpticsActual
An object is placed at an unknown distance from a convex objective lens of focal length 8 cm. The objective lens forms a real image which acts as an object for a convex eyepiece of focal length 6.25 cm. The distance between the objective and eyepiece is 45 cm. The microscope is adjusted so that the final image is formed at the least distance of distinct vision (25 cm). Which of the following is correct?
Options
- AObject distance = 7.5 cm; Total magnification = 10
- BObject distance = 10 cm; Total magnification = 20
- CObject distance = 5 cm; Total magnification = 20
- DObject distance = 2.5 cm; Total magnification = 10
Correct answer
B. Object distance = 10 cm; Total magnification = 20
Step-by-step solution
For the eyepiece, the final image is formed at the least distance of distinct vision, so v_e = -25 cm . Using the lens formula for the eyepiece: 1 v_e - 1 u_e = 1 f_e 1 -25 - 1 u_e = 1 6.25 - 1 u_e = 1 6.25 + 1 25 = 4 25 + 1 25 = 5 25 = 1 5 u_e = -5 cm The distance between the objective and the eyepiece is L = v_o + |u_e| , where v_o is the image distance for the objective lens. 45 = v_o + |-5| v_o = 45 - 5 = 40 cm Using the lens formula for the objective lens: 1 v_o - 1 u_o = 1 f_o 1 40 - 1 u_o = 1 8 - 1 u_o = 1 8