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A converging lens of focal length fconverging = 26.6 cm is now inserted at x = x3 = -52.86 cm.  In the absence of the diverging lens, at what x co-ordinate, x4, would the image of the arrow form?

3) A converging lens of focal length fconverging - 26.6 cm is now inserted at x = x3 =-52.86 cm. In the absence of the diverging lens, at what x co-ordinate, X4, would the image of the arrow form? light, image in absence of diverging lens 4 3 cm Submit 4) To determine the image of the arrow from the combined converging + diverging lens system, we take the image from the converging lens (in the absence of the diverging lens) to be the object for the diverging lens. If this image is downstream of the diverging lens, the object for the diverging lens is virtual All this means is that the rays entering the diverging lens are converging towards a location downstream of the diverging lens. The final image can be calculated using this virtual object distance and the focal length of the diverging lens. If this description seems unclear to you, you can check out the more detailed description of multiple lens systems given in the Optical Instruments unit. What is x5, the x co-ordinate of the final image of the combined system? cm Submit 5) Is the final image of the arrow real or virtual? Is it upright or inverted? Real and upright Real and inverted Virtual and upright Virtual and inverted Submit

Explanation / Answer

1/u - 1/v = 1/f
v = -33.9cm
f = -52.9cm
u = -20.66 cm
m = h'/h = v/u
h' = 10.993 cm

1) u = -20.66 cm
2) h' = 10.993 cm
3) u = -20.66 + 52.86 = 32.2 cm
f = -26.6 cm
v = 14.566 cm
4) u = 14.566 cm
f = 52.9 cm
v = 20.102 cm

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