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 = 9.7 cm. The second lens is located at x = x2 = 53.9 cm and has a focal length of f2 = 18 cm. The tip of the object arrow is located at (x,y) = (xo,yo) = (-14.6 cm, 6 cm).

1)

What is x1, the x co-ordinate of image of the arrow formed by the first lens?

cm

Your submissions:

2)

What is y1, the y co-ordinate of the image of the tip of the arrow formed by the first lens?

cm

Your submissions:

3)

What is x3, the x co-ordinate of image of the arrow formed by the two lens system?

cm

Your submissions:

4)

What is y3, the y co-ordinate of the image of the tip of the arrow formed by the two lens system?

cm

Your submissions:

5)

What is the nature of the final image relative to the object?

Real and Inverted

Real and Upright

Virtual and Inverted

Virtual and Upright

Your submissions:

6)

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 = -9.75 cm, keeping the second lens at x = 53.9 cm.

Move the second lens to x = 62.9 cm, keeping the first lens at x = 0.

Move the second lens to x = 37.9020408163265 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

1) 1/v=1/f+1/u=1/9.7+1/-14.6

so v=28.9cm

2) v/u=h2/h1

so 28.9/-14.6=h2/6

so, h2=-11.88cm

3) 1/v=1/f+1/u

so 1/v=1/18+1/(28.9-53.9) because object distance u is from the mirror

so v=64.28cm

so x3=64.28+53.9=118.18cm

4)v/u=h2/h1

so 64.28/-25=h2/-11.88

so h2=30.545cm

5) Real and upright

6) Move the second lens to x = 37.9020408163265 cm, keeping the first lens at x = 0

because then the vaule of u will be less then value of focal length of the lens

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