An object 1.2 cm high is placed 21.7 cm to the left of a converging lens of foca

Question

An object 1.2 cm high is placed 21.7 cm to the left of a converging lens of focal length 14.2cm. (a). Find the position of the image of this object relative to this lens. Use a positive sign if the image lies to the right of the lens, and vice-versa. (b). A second converging lens of focal length of 14.3 cm is placed 14.0 cm to the right of the first lens. Find this object distance, i.e. the distance between the first image and lens 2. (c). How far from the second lens is the image that it forms? (d). What is the height of the final image? I already got the answer to part (a). which is 41.09 cm.

Explanation / Answer

m = magnification Real is Positive sign convention. Converging Lens: 1 / 20 + 1 / v = 1 / 25 v = - 100 cm. m = - 100 / 20 = - 5.0 Virtual image 100 cm. left of the converging lens. Magnification 5.0. Virtual image becomes a real object 125 cm. left of the diverging lens. 1 / 125 + 1 / v = - 1 / 10 v = - 9.3 cm. m = - 9.3 / 125 Virtual image 9.3 cm. left of the diverging lens. Magnification = 5.0 * 9.3 / 125 = 0.370.

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