The figure below shows an object placed a distance d_oi from one of two convergi

Question

The figure below shows an object placed a distance d_oi from one of two converging lenses separated by s = 1.00 m. The first lens has focal length f_1 = 21.0 cm, and the second lens has focal length f_2 = 50.0 cm. An Image is formed by light passing through both lenses at a distance d_12 = 16.0 cm to the left of the second lens. (Include the sign of the value in your answers.) What is the value of d_01 that will result in this image position? cm Is the final image formed by the two lenses real or virtual? real virtual What is the magnification of the final image? Is the final image upright or inverted? upright inverted

Explanation / Answer

a) For second lens:

1/do2 + 1/di2 = 1/f2

=> 1/do2 + 1/(-16) = 1/50

=> do2 = 50 * 16 / (50 + 16) = 12.1 cm

So, di1 = (100 - 12.1) cm = 87.9 cm

For first lens:

1/do1 + 1/di1 = 1/f1

=> 1/do1 + 1/87.9 = 1/21

=> do1 = 87.9 * 21 / (87.9 - 21) = 27.6 cm

b) The image is virtual since the object-distance for the second lens is positive and the image distance negative.

c) m = m1m2 = (-di1/do1) * (-di2/do2) = di1di2/do1do2 = 87.9 * (-16) / (27.6 * 12.1) = -4.2

d) Since the overall magnification is negative, image is inverted.

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