Question

Problem 33.77 Part A Constants I Periodic Table A microscope has a 2.1 -em-focal-length eyepiece and a 0.90 -cm objective. Assuming a relaxed normal eye Calculate the position of the object if the distance between the lenses is 17.1 cm Express your answer using two significant figures d11.9 cTm Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part B Caiculate the total magnification Express your answer using two significant figures Submit Previous Answers Reauest Answer Incorrect; Try Again;4 attempts remaining

Answer 1

Given, The focal length of eye piece(f_c) = 2.1 cm [f_00 objective = 0.90 cm Distance between the lenses(L) = 17.1 cm (a) We know, in case of a micro scope image from the objective is at the focal length of the eye piece. So, the image distance for the objective is d_10 = L - f_e = 17.1 cm - 2.1 cm = 15 cm The Lens� equation for the objective is (1/d_00) + (1/d_10) = 1/f_0 (d_00 = object distance) (1/d_00) + (1/15) = 1/0.90 1/d_00 = 1/0.90 - 1/15 = 10/9 - 1/15 = 141/135 d_00 = 135/145 = 0.95 cm s0, object distance is 0.95 cm b) For relaxed normal eye, final image is at infinity. so, magnification of eye piece is M_e = N/f_e = 25cm/2.1 cm = 11.90 magnification of objective m_0 = d_10/d_00 = 15cm/0.95cm = 15.78 Total magnification m = m_em_0 = 11.90 times 15.78 M = 187.78