Question

A diverging lens located in the y-z plane at x - 0 forms an image of an arrow at x = x2 =-27.5 cm. The image of the tip of the arrow is located at y -y2 - 5 cm. The magnitude of the focal length of the diverging lens is 49.1 cm light (x2.y2) image 1) What is x1, the x co-ordinate of the object arrow?. 62.5 You currently have 7 submissions for this question. Only 15 submission are allowed You can make 8 more submissions for this question cm Submit 2) What is y1, the y co-ordinate of the tip of the object arrow? 11.36 cm Submit 3) A converging lens of focal length fconverging 16.06 cm is now inserted at x X3-41.26 cm. In the absence of the diverging lens, at what x co-ordinate, X4, would the image of the arrow form? light mage in absence of diverging lens 4 9.15 cm Submit

# 1 and 2 are already done correctly, please do not solve

Answer 1

The object distance from the converging lens is. u = -(62.5 - 41.26) u = -21.24 cm The distance of the image formed by the converging lens in the basence of the diverging lens is V = fu/ft u = 16.06 times (-21.24)/16.06 - 21.24 = -341.1144/-5.18 V = 65.85 cm The x-coordinate of the image formed by the converging lens is x_4 = 65.85 - 41.26 x_4 = 24.59 cm The image formed by the converging lens acts as object for the diverging lens the object distance for the diverging lens is u' = 24.59 cm The distance of final image formed by the diverging lens is V = fu'/ftu' = -49.1(24.59)/-49.1 - 24.59 V = -1207.369/-73.69 = > 16.384 cm The final image is real and inverted.