Question

e), f): i, ii

We have seen that (at least for the above example) that we can treat the image of one lens as the object for the next. Also, the overall magnification can be found by combining the individual magnifications. So let's investigate what happens when this first image falls at various positions relative to the second lens and what different types of lenses do to the final image. Begin by setting up the optical bench as follows Set up the optical bench lamp with the letter F at one end of the bench. Place an f -+10 cm lens at a distance of 30 cm from object. e) Locate Image #1 with a screen (Make a note of its position). Record the object and image distances and compute the magnification for the FIRST lens. The bench lamp and the first lens will remain in the above position for each of the four examples below. These are just examples and are in no means all the possibilities f) Place a second lens f; - +10 em at a distance 30 cm PAST the location of Image #1. 1) Locate Image #2. Record the object and image distances and compute the magnification for the SECOND lens. ii) Is Image #2 bigger, same size, or smaller than Object #1? Inverted or same orientation as Object #1? Is this consistent with the computed magnification Mor - MM? The techniques for lenses work in all cases as long as we understand and use the proper sign conventions for the object and image distances. Consider what happens if we place a second lens BEFORE the location of Image #1. That is to say, consider light that passes through lens #1. It is heading toward forming an image at the location of Image #1, but before it can form Image #1, it is bent again by lens #2. This is what's known as a virtual object. We can still use Image #1 - Object #2 and apply all the rules to this case as long as the signs are handled correctly. A real object is located in front of a lens on the incoming light side. We have assigned this as a positive object distance. A virtual object would have formed behind a lens on the outgoing light side. Thus by extension of our convention, this is assigned a negative object distance. 2 235

Answer 1

e) When the object is placed at 30cm in front of a f=10cm lens, i.e. u=-30cm , and f=10cm,

then, using lens formula,

t '1 30 10 15

so, image#1 is at a distance of 15 cm from the first lens. i.e. v₁=15cm

Now, magnification m1=v1/u =15/(-30) = -(1/2)

So the image will be diminished and inverted.

f)Now, a lens is placed 30cm past the image#1,

So, image#1 acts as an object for lens 2.

i.e in this case again, u=-30cm, f=10cm

So,

$\frac{1}{v_{2}}=\frac{1}{u}+\frac{1}{f}=\frac{-1}{30}+\frac{1}{10}=\frac{1}{15}$

So, image#2 will be formed at 15cm from the second lens.

magnification m2=v₂/u=15/(-30)=-(1/2)

So, the image will again be inverted and diminished.

Now, the resultant magnification M=m1*m2 = (1/4)

So the final image will be diminished 4 times but will be straight, i.e. not inverted wrt to object.