For a compound microscope the magnifying power (MP) is the objective magnification (m) multiplied by the eyepiece magnification (Me). Generally two small focal length lenses are used with around the same value of f. A simple magnifier has been set up with a virtual image produced at 25 cm away from the lens of focal length fe. The object is placed at the required distance 7.2 cm away and can be viewed. Another lens is now added as the objective in a compound microscope. The first lens is the eyepiece. The objective lens has focal length of fo = 9.5 cm and for a magnification of m = -2 it can be calculated that the object now needs to be placed 14.25 cm away from this lens. How far apart must the two lenses be set to construct a compound microscope with the virtual image produced at 25 cm?Give your answer to 4 significant figures.

To solve this problem, we need to use the lens formula and the magnification formula.

The lens formula is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

The magnification formula is m = -v/u.

First, we need to find the image distance (v) for the objective lens. We know that the magnification (m) is -2 and the object distance (u) is -14.25 cm (negative because the object is on the same side of the lens as the light source).

We can rearrange the magnification formula to solve for v: v = mu. Substituting the given values, we get v = -2 * -14.25 cm = 28.5 cm.

Next, we use the lens formula to find the image distance (v) for the eyepiece. We know that the focal length (f) is 9.5 cm and the object distance (u) is -28.5 cm (negative because the object is on the same side of the lens as the light source).

Rearranging the lens formula to solve for v, we get v = 1/(1/f + 1/u). Substituting the given values, we get v = 1/(1/9.5 cm + 1/-28.5 cm) = -15.789 cm.

Finally, the distance between the two lenses is the sum of the absolute values of the image distance for the objective lens and the image distance for the eyepiece. This gives us |28.5 cm| + |-15.789 cm| = 44.289 cm.

So, the two lenses must be set 44.289 cm apart to construct a compound microscope with the virtual image produced at 25 cm. This answer is given to 4 significant figures.