A 6.47-mm-high firefly sits on the axis of, and 13.1 cm in front of, the thin lens A, whose focal length is 6.19 cm. Behind lens A there is another thin lens, lens B, with focal length 27.9 cm. The two lenses share a common axis and are 55.7 cm apart. Is the image of the firefly that lens B forms real or virtual? How far from lens B is this image located (expressed as a positive number)? What is the height of this image (as a positive number)? Is this image upright or inverted with respect to the firefly?

Explanation:

For lens A

object distance u = - 13.1 cm , focal length f = 6.19 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/13.1 = 1/6.19

1/v = 1/6.19 - 1/13.1

= .16155 - .07633

= .08522

v = 11.7 3 cm

For  lens B

object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/43.97 = 1/27.9

1/v = 1/27.9 - 1/43.97

= .03584 - .022742

= .013098

v = 76.35 cm

Image will be formed 76.35 cm behind lens B .

magnification of lens system

= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B

= (11.73 / 13.1) x (76.35 / 43.97)

= .8954 x 1.73

= 1.5547

size of image = total magnification x size of object

= 1.5547 x 6.47

= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.

answer: strength i guess

explanation: