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.

e) none of the above answers are correct

need to see circuit to get time constant