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 6.3 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 calculate the image distance (v) for the eyepiece using the lens formula. We know that the object is placed at a distance of 6.3 cm from the eyepiece (u = -6.3 cm, since the object is on the same side of the lens as the light is coming from), and the focal length of the eyepiece (fe) is 25 cm.

1/fe = 1/v - 1/u 1/25 = 1/v - 1/-6.3 1/25 + 1/6.3 = 1/v v = -15.789 cm (to 4 significant figures)

This means that the image formed by the eyepiece is 15.789 cm on the other side of the lens.

Next, we calculate the object distance for the objective lens. We know that the magnification of the objective lens (m) is -2, and the focal length of the objective lens (fo) is 9.5 cm.

Using the magnification formula, we get: m = -v/u -2 = -15.789/u u = 15.789/2 u = 7.894 cm (to 4 significant figures)

This means that the object for the objective lens must be placed 7.894 cm from the lens.

Finally, we calculate the distance between the two lenses. This is the sum of the image distance for the eyepiece and the object distance for the objective lens.

Distance = 15.789 cm + 7.894 cm = 23.683 cm (to 4 significant figures)

So, the two lenses must be set 23.683 cm apart to construct a compound microscope with the virtual image produced at 25 cm.