Consider the image formed by a thin diverging lens. Decide if the image will always, sometimes, or never meet?

Consider the image formed by a thin diverging lens. Decide if the image will always, sometimes, or never meet each of the following conditions.


(a) inverted


---Select--- never always sometimes





(b) upright


---Select--- never always sometimes





(c) real


---Select--- never always sometimes





(d) virtual


---Select--- never always sometimes





(e) larger than the object


---Select--- never always sometimes





(f) smaller than the object


---Select--- never always sometimes








PLEASE HELP QUICK!!

Consider the image formed by a thin diverging lens. Decide if the image will always, sometimes, or never meet?
A thin diverging lens by itself will always cause the light rays coming from a single point to... you guessed it, diverge.





The rays coming from the single point are already diverging from the point before they reach the lense, they then diverge further. The light rays will never again actually pass through a single focusses point. Thus the image will always be virtual, appearing to come from a point between the object and lens. It will also always appear smaller and upright.





You can prove visually by following the rules for lens diagrams. For a diverging lens, light coming from the object through the centre of the lens will continue straight. Light coming from the object parallel to the horizontal axis will diverge to appear to come from the focal point on the object side of the lens. Light that was originally heading towards the focal point on the opposite side of the lens will now travel horizontally. All these 3 lines will converge on the object side of the lens. That is, the light appears to come from a virtual object.





Now, if you draw this diagram with the object behind, on or in front of the focal point, the virtual image is always upright and smaller.





Also note that if you rearrange the thin lens equation to give


s2 = 1 / ( 1/f - 1/s1)


a negative focal length will always give you a negative s2, proving that the image is on the same side as the object.


The magnification


M = f / (f - s1)


for a negative f will always be positive and less than one, hence the image is upright and smaller.





Let f = -a (a always positive, f always negative), then


M = -a / (-a - s1)


M = a / (a + s1)


Since a and s1 are always positive, M must be positive


a + s1 must always be greater than a


so a / (a + s1) must be less than 1