A converging lens of focal length 8.060cm is 20.5cm to the left of a diverging l

Question

A converging lens of focal length 8.060cm is 20.5cm to the left of a diverging lens of focal length -6.25cm . A coin is placed 12.5cm to the left of the converging lens. PART A Find the location of the coin's final image. PART B Find the magnification of the coin's final image.

Explanation / Answer

A converging lens of focal length 8.04 cm is 24.3 cm to the left of a diverging lens of focal length -6.06 cm. A coin is placed 11.4 cm to the left of the converging lens. Calculate the location of the coin's final image. Calculate final magnification The image of the first lens becomes the object of the second So 1/s'1 + 1/s1 = 1/f.1.....=>1/s'1 = 1/f 1- 1/s1 = 1/8.04 - 1/11.4 = 0.03666 Therefore s'1 = 27.3cm This means the object for the second lens is 27.3 - 24.3 = 3.0 cm behind the diverging lens...so s = -3.0 (a virtual object) Now for the second lens we have 1/s'2 + 1/s2 = 1/f2 So 1/s'2 = 1/f 2- 1/s2 = 1/(-6.06) - 1/(-3.0) = 0.1683 So s'2 = 5.94cm So the image forms 5.94 cm to the right of the diverging lens Final magnification = m1*m2 = s'1/s1*s'2/s2 = 27.3/11.4*5.94/3 = 4.74

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