Question

part D. The answer is not 60.4.

(10%) Problem 7: Two lenses are mounted d 28 cm apart on an optical bench. The focal length of the first lens ish = 8.7 cm and that of the second lens isf2 = 4.4 cm. An object of height ho-4.4 cm is placed at a distance of do-22 cm in front of the first lens 25% Part (a) Ignoring the second lens for now, at what distance, in centimeters, behind the first lens is the object's image formed by that lens? d,-14.39 Correct! 25% Part (b) Calculate the magnification of that image, including its sign m0.654 m=-0.654 V Correct! Y 25% Part (c) Which choice best describes that image? Real and inverted. Correct! ** 25% Part (d) Now consider the two-lens system and the final image it forms, ie., the image created by the second lens. What is the distance, in centimeters, between the object and its final image? Grade Summary Deductions Potential D- 60l 28% 72% 789HOME Submissions sinO cotan0 asi cosO Attempts remaining: 995 (4% per attempt) detailed view acos atan acotan0 sinh coshO tanh0 cotanh0 Degrees ORadians 4% 4% 4% 4% 0 END BACKSPACE | DEL | CLEAR 4. Submit I give up!

Answer 1

For first lens. V_1 = ? u_1 = minus 22 cm, f_1 = 8.7 Lens formula rightarrow 1/v_1 minus 1/u_1 = 1/f_1 = 1/v + 1/22 = 1/8.7 1/v_1 = 1/8.8 = 1/22 v_1 = 14.39 cm. Now, for lens u_2 = minus(28 minus 14.39) = minus 13.61 f_2 = 4.4 cm. so, 1/v^2 minus 1/u^2 = 1/f^2 1/v^2 = 1/4.4 minus 1/13.61 v_2 = 6.50 cm. So, Total distance = 22 + 28 + 6.5 = 56.5 cm