A thinner biconvex lens has a longer focal length f, and bends light rays through a smaller angle.
If object distance u > f, the real image formed is further away from the lens (larger image distance v) and larger, compared to case with a smaller f.
See https://www.dropbox.com/s/m87kdw4f4f1klyw/Photo%2011-1-14%207%2040%2020%20pm.jpg

WS 8.3 Q1(d)
Problem solving strategy: Draw additional rays that are parallel to given ray; draw ray passing through optical centre of lens.
- parallel rays (not horizontal) will meet (intersect) at a point on the focal plane (a vertical plane at the focal point).
- ray passing through centre of lens would go straight through; so this ray meets the focal plane at a point (called it P) and hence all other parallel rays would also pass through P.

WS 8.3 Q1(a)
When horizontal light rays (e.g. from distant object) pass through a lens, they would converge and pass through the focal point F. Likewise in a reverse direction, any light rays passing through a focal point F will then become horizontal after passing through the lens (the 3rd type of standard path of light rays, see page 271 in textbook).

(Note: The above idea is also called the principle of reversibility of light.)
Ref: Basic http://schools.matter.org.uk/SchoolsGlossary/reversibility.html
Ref: Advance: http://electron6.phys.utk.edu/optics421/modules/m1/Fermat's%20principle.htm

WS 8.2 Q2(f): u = infinity
The object is at infinity (very far away), so we can't draw the object on the small scale, but we draw parallel light rays from it.
Such a light source is the same as what you used (physics lab window grilles far away) in practical PR1 to estimate the focal length of a lens.

The eye only sees one set of light rays that reach its retina. These light rays (converging at the retina to give a sharp image) appear to have come from an "object" which upright, magnified and behind the original object - this is what the eye sees as the image.
The light rays from actual object no longer appear to come from its original position - this object is no longer seen by the eye.
The effect of this converging lens (only see magnified virtual image), is different from that of a transparent plane glass (which has little effect on the original light rays).

Problem-solving strategy:
1. For each ray (A), we draw another ray B parallel to it and passing through the optical centre, so it goes straight through.
2. Parallel rays will converge (meet) at the focal plane (vertical plane at F), and this point is given by ray B intersecting the focal plane (at point say C).
3. Now we can draw the original ray A after the lens, also passing through the same point C.

Wrt the ray diagram, if the object is to the left of the lens,
1. we expect a real image to the right of the lens, so we would put a screen on the right of the lens (we just draw the real image, not the eye), and view the screen where the light rays converge, just like your practical PR 1.
2. We expect a virtual image to the left of the object, so we would position our eyes to the right of the lens and view the virtual image, as shown in http://www.leydenscience.org/physics/electmag/con_inf.jpg

AS 8.2:
Q3: Refer to a ray diagram with triangles formed by a ray through the optical centre of lens.
Together with the vertical lines (of object & image), the ray gives similar triangles on both sides of the lens.
The object distance u & image distance v, both are proportional to the height of object/image.
(a)(i) real image: image is 3 times object also means v/u = 3
So f given, v = 3u, substitute into lens formula to solve for u.
(a)(ii) virtual image: v = -3u, similarly, find u.

(b) reduce size of image means reduce reduce v. This can be done by increasing u.

AS 8.2, Q2: Using the lens formula requires a sign convention:
If image is real v is positive, if image is virtual v is negative.
Since image is on the same side as the object, it is virtual, so v = -150 mm.
Apply the lens formula to find f.

Negative sign was used: v = -50.0 must be substituted into the lens formula.

1. Yes, one for v = + infinity and - infinity
2. If N is a non-zero number: N / infinity = 0 and infinity / N = infinity
3. u = infinity means object at infinity, we draw oblique (at an angle) parallel rays incident on the lens (including one ray through optical centre), so that rays meet at a point on the focal plane, and a real, inverted and diminished image can be drawn.

The rule of thumb, u increases as v decreases and vice versa is only true for real inverted images when u > f (object on one side, image on other side of lens).
This includes f < u < 2f.
The rule is no longer valid when
a. u = f when image at v= +/- infinity
b. u < f when image is virtual, and v is negative
Also refer to the graph of v against u plotted at https://www.dropbox.com/s/562h8k1xdqbh2nm/Photo%208-5-14%201%2018%2052%20pm.jpg

What do you think is(are) possible answer(s)? Explain your reasoning.

Celeste: Is the answer number (2), because as the object distance becomes further, the image distance gets smaller, hence the point of convergence of the light rays falls on the screen, forming the sharpest image?