A lens of focal length 150 cm is 10 cm to the left of a seco

A lens of focal length +15.0 cm is 10 cm to the left of a second lens of focal length -15.0 cm. Where is the final image of an object that is 30 cm to the left of the positive lens? The image formed by the lens system is located A lens (f_A = 15.0 cm) is located 10 cm to the left of a second lens (f_B = -15.0 cm) with the same optical axis. An object is placed a distance d_O, A = 30 cm in front of the first lens. We can use the thin-lens equation to determine the location of the image produced by lens A. This image now acts as the object for the second lens (lens 3), sc we can use the thin-lens equation again to calculate the location of the image after it passes through both lenses, cm in front of the second lens. Is the image real or virtual? The final image is virtual because d_I, B 0. The final image is real because d_I, B 0. How do the image\'s size and orientation compare to those of the original object? Explain your answer. Whether or not the image is upright or inverted can be found by calculating the total magnification due to both lenses. (d) How must you be oriented to see the image? To see the final image, a viewer would need to look to the right through both lenses. To see the final image, a viewer would need to look to the left through both lenses. To see the final image, a viewer would need to look to the right through the first lens. To see the final image, a viewer would need to look to the left through the first lens.

Solution

Answer:

Image for the negative lense when the object is at 30 -10 = 20 cm

1/f = 1/v + 1/u

1/(-15) = 1/v + 1/(-20) (Taking positive sign towards right of lense)

1/v = 1/15 or v = - 60 cm

now distance of object for convex lense = 60 + 10 =70

1/(-15) = 1/(-70) + 1/v\' (Taking positive sign towards right of lense)

v\' = - 19 cm

Total magnification = m1 x m2