A system of two lenses forms an image of an arrow at x = x 3 = 69 cm. The first

Question

A system of two lenses forms an image of an arrow at x = x3 = 69 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f1 = 9.9 cm. The second lens is located at x = x2 = 29.6 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (xo, yo) = (-42 cm, 18.1 cm).

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

-8 cm

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

3.45 cm

3) What is f2, the focal length of the second lens. If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative.

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

5) The positions of the two lenses are now interchnaged (i.e., the second lens is moved to x = 0 and the diverging lens is moved to x = x2 = 29.6 cm). What is the nature of the final image in this new system?

Explanation / Answer

1)for the first lens,

1/u +1/v=1/f

or 1/42 +1/v=-1/9.9

or v=-8.011

so x=-8.011

2)m=v/u

=8.011/42

=0.1907

so y'=0.1907*18.1

=+3.45 cm

3)u=8.011+29.6

=37.611 cm

v=69-29.6

=39.4 cm

so,

1/u +1/v=1/x

or 1/37.611 + 1/39.4=1/f

or f=19.2424 cm

the lens is converging with f2=19.2424 cm

4)for the second lens, image will be inverted.

so m=v/u

=39.4/37.611

=1.047

so y'=-my

=-1.047*3.45

=-3.612 cm

5)for the first lens,

1/42+1/v=1/19.2424

or v=35.5136

for the second lens,

u=29.6-35.5136

=-5.9136 cm

so,

-1/5.9136 + 1/v=-1/9.9

or v=14.68 cm

so here we come to know that for the second refraction,

v/u=-ve

so the image is virtual and inverted

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