Physics help please A converging lens with a focal length of

A converging lens with a focal length of 12.0 cm forms a virtual image 7.50 mm tall, 18.0 cm to the right of the lens. Determine the position of the object. (Give the object distance with the correct sign.) cm Determine the size of the object. mm Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? opposite sides same side

Solution

Use the equation

1/si + 1/so = 1/f

where si = -18 cm and f = 12. si is negative because you stated that the image is virtual. With these two values, solve for the equation so above. You should get 7.2 cm . Once you have this value, you can calculate the size of the object using this equation

Magnification = di / do = si / so

then

do = di so / si =7.5 * 18/7.2 = -18.75

Virtual images in positive (converging) lenses are always erect and are always on the same side of the lens as the object is.