A system of two converging lenses forms an image of an arrow as shown. The first

Question

A system of two converging lenses forms an image of an arrow as shown. The first lens is located at x = 0 and has a focal length of f1 = 6.5 cm. The second lens is located at x = X2 = 45.45 cm and has a focal length of f2 = 16.7 cm. The tip of the object arrow is located at (x,y) = (X0,y0) = (-9.1 cm, 5.6 cm). What is x1, the x co-ordinate of image of the arrow formed by the first lens? What is y1, the y co-ordinate of the image of the tip of the arrow formed by the first lens? What is X3, the x co-ordinate of image of the arrow formed by the two lens system? What is y3, the y co-ordinate of the image of the tip of the arrow formed by the two lens system? What is the nature of the final image relative to the object? Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright Which of the following changes to the locations of the lenses would result in a virtual and inverted image of the original object arrow? Move the first lens to x = -5.85 cm, keeping the second lens at x = 45.45 cm. Move the second lens to x = 53.8 cm, keeping the first lens at x = 0. Move the second lens to x = 31.1 cm, keeping the first lens at x = 0. None of the above moves will produce a virtual inverted image of the object arrow.

Explanation / Answer

For first lens

u = -9.1 cm

f = 6.5 cm

1/v - 1/u = 1/f

So, v = 22.75 cm

M = v/u = -2.5

hi = M*ho = -14

Now, for second lens...

the object is at (45.45 - 22.75) cm from lens...

so, u = -22.7 cm

f = 16.7 cm

1/v - 1/u = 1/f

so, v = 63.18 cm

M = v/u = -2.78

hi = M*ho = -2.78 * 14 = 38.967 cm

So answers :

1) 22.75 cm

2) -14 cm

3) (63.18+45.45) = 108.63 cm

4) 39.967 cm

5) Real and Upright

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