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1. The beam of light will travel down the fiber if total internal reflection takes place at the interface between the core and the cladding. The critical angle is given by

Total internal reflection will take place if , i.e. if

Applying Snell’s Law at the air/glass interface, we get

Now

Therefore Equation (1.2) may be written as

whence

(1.7)

2. For a converging (concave) mirror, the coordinates u and v (we are using the Cartesian sign convention) are related to the coordinates by the transformation equations

Assuming that the light impinges on the mirror from the left (principal focus on the left of the mirror: ), the mirror formula may therefore be written in the form

i.e.

(2.4)

For a diverging (convex) mirror, the principal focus is located on the right of the mirror ( ), assuming that the light impinges on the mirror from the left. The transformation equations are now

Thus the mirror equation becomes

which is the same as Equation (2.2), with F replaced by . The Newtonian form (2.4) therefore applies to a diverging mirror as well.

Since is positive, it follows from Equation (2.4) that the coordinates must be of the same sign. In other words, the object and the image must be on the same side of the principal focus (we assume that the object is not located at the principal focus or at infinity). Note, further, that Equation(2.4) may be written in the form

where are the distances of the object and the image, respectively, from the principal focus. Since F is fixed, decreases as increases, and vice versa. This means that an object and its image in a spherical mirror (converging or diverging) move in opposite directions.

3. Using the Cartesian sign convention, the lens formulae may be written as

where u, v and f are the Cartesian coordinates of the object, image and principal focus, respectively (relative to the pole of the lens), and m is the magnification. Let us assume that the object is located on the left of the lens ( , where U is the distance of the object from the lens), i.e. light impinges on the lens from the left. For a diverging lens, parallel incident rays diverge, so the principal focus will be on the left of the lens ( , where F is the focal length of the lens). The lens formula then gives

and we see that the image coordinate v is negative. This means that the image is located on the left of the lens. Now, a real image, which is produced by converging rays, would be located on the right of the lens. If the image is on the left of the lens, the refracted rays must diverge, producing a virtual image. Thus we see that a diverging lens cannot produce a real image of a real object.

A real image can, however, be formed by a diverging lens when the incident rays are converging, producing a virtual object O a distance U to the right of the lens (we again assume that the light impinges from the left). Substituting in the lens formula, we get

It follows that if , the image coordinate v will be positive ( ), i.e. the final image I will be located on the right of the lens. Note that

This shows that the refracted rays converge (more slowly than the incident rays) to produce a real image on the right of the lens.

Using the Cartesian sign convention, the lens magnification is given by

assuming, as always, that the object is on the left of the lens ( ). If the refracted rays converge, a real image will be produced on the right of the lens ( ), and the magnification becomes , which is negative, implying that the image is inverted. If, on the other hand, the refracted rays diverge, a virtual image will be produced on the left of the lens ( ), and the magnification now becomes , which is positive, implying that the image is upright.

For the first (converging) lens, we may write

Substituting the numerical values and , we get

The image formed by the converging lens serves as the object for the diverging lens. It is located a distance of to the right of the diverging lens ( ). However, since the light impinges on the diverging lens from the left, its principal focus is on the left ( ). The position and nature of the final image are given by

The final image I is located a distance of 30 cm to the left of the diverging lens. The overall magnification is obtained from the relation

which implies that the final image is upright, and the same size as the object.