A camera. Where is the film with respect to the focal point?

Imagine a camera with a single lens with a focal distance f = 35 mm. By how much and in what direction should the lens be moved to move its focus from an object, which is far-far away to one at a distance of 1.5 m? Far-far away means l = , and l’ = f = 35 mm. New distance l = 1.5 m ; l’- ?

A movie projector. Where is the film with respect to the focal point?

The diameter of an eyeball is about 23 mm. It focuses light emanating from different objects onto the retina to produce sharp images. Most of the focusing job is done by the cornea, which acts as a fixed lens with a focal distance of 23 mm. The lens is only doing some fine tuning to move the focusing from objects which are far-far away to objects which are close.

By changing the focal distance of the lens! We always have l’  23 mm. Far-far away means l = , and f = l ’  23 mm. A healthy human eye can clearly see (focus) objects at distances from infinity to about 25 cm. How is that achieved? An object at l = 25 cm means The adaptive lens driven by the eye muscles changes the focal distance of the eye by “only” 10%. But this is quite a lot!..

Accommodation from infinity to about 25 cm. How much does the “strength” of the lens change? When the eye accommodates from infinity to 25, its refractive power increases by It is equivalent to putting in front of the eye a lens with refractive power of 4 dpt and a focal length of The strength of a lens is called refractive power. It is measured in diopters. f is in meters.

Myopic (shortsighted) eye – the lens is always “too strong”, that is too much converging and if the object is far away it creates its image in front of the retina. So, the light pattern on the retina becomes blurry, out of focus. The eye can be helped by a negative, diverging lens, which creates virtual images of far away objects closer to the eye. Rephrasing it: the diverging lens + the “too much” converging lens in the eye make a composite lens of the right converging power.

Hyperopic (farsighted eye) – the lens is sometimes not quite strong enough. If an object is close it focuses behind the retina. So, the light pattern on the retina becomes blurry, out of focus. The eye can be helped by a positive, converging lens, which creates magnified virtual images of close objects further away from the eye. Rephrasing it: the additional converging lens + the “not strong enough” converging lens in the eye make a composite lens of the right converging power.

Optical instruments: magnifying glasses, spyglasses, telescopes… What do they do?   The mother elephant looks larger than the cub elephant, because the mother subtends larger angle,  >  , and has large angular size. Therefore, we can better see mother’s ear than cub’s ear. The situation changes if the cub is closer to the observer.  A spyglass creates a virtual image, which is closer to the observer and has a larger angular dimension. So, one can better see the details

Magnifying glass. Your eye cannot see objects clearly from closer than the “near point”, L = 25 cm. So, the largest angular size of a fine-print object with height h is  = h/L. A magnifying glass creates a virtual image with the same angular size as the object, but you can now have the object at a small distance f from your eye. So, the angular size is  = h/f. Magnification of the magnifying glass:

Two sources oscillating in phase. Lines of nodes and antinodes? Are antinodes the points, where disturbance is always high? Are nodes the point, where disturbance is always low? If you go along the green line (upward), you will see a pattern of nodes and antinodes like in a standing wave.