A 2.8-cm-tall object is located 3.3 cm to the left of a converging lens with a f

Question

A 2.8-cm-tall object is located 3.3 cm to the left of a converging lens with a focal length of 4.3 cm. A diverging lens, of focal length -6.0 cm, is 13 cm to the right of the first lens.
Find the position of the final image.
Find the size of the final image.
Find the orientation of the final image.

Explanation / Answer

for the first case using lensmakers equations 1/o + 1/i = 1/f we take the sign of the distances according to the convention so, 1/-3.3 + 1/x = 1/4.3 so calculating x = 1.87 cm magnification = i/o 1.87/-3.3 = -0.5657 I_h / O_h = I_h /2.8 = -0.5657 => I_h = 1.583 cm and will be inverted this becomes the object for the other lense so, now using the equation for the other lense o = 13- 1.87 = 11.13 1/-11.13 + 1/x = -1/6 x = -13.017 cm from the second lense i / o = -13.017/-11.13 = 1.1695 image height = 1.583 * 1.1695 = 1.85131 and will be errect If you have any further querry then do let me know!

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