Question 19M.3.SL.TZ2.11
| Date | May 2019 | Marks available | [Maximum mark: 10] | Reference code | 19M.3.SL.TZ2.11 |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Calculate, Construct, Determine, Identify, Label, Outline, Suggest | Question number | 11 | Adapted from | N/A |
A student places an object 5.0 cm from a converging lens of focal length 10.0 cm.
Construct rays, on the diagram, to locate the image of this object formed by the lens. Label this with the letter I.
[2]
any two correct rays with extensions ✔
extensions converging to locate an upward virtual image labelled I with position within shaded region around focal point on diagram ✔
The simple ray diagram was constructed well by most candidates, especially compared to previous years.
Determine, by calculation, the linear magnification produced in the above diagram.
[2]
v = «–» 10«cm» ✔
M «= –=–» = «+» 2 ✔
The very simple calculation of magnification was done well by nearly everybody.
Suggest an application for the lens used in this way.
[1]
magnifying glass
OR
Simple microscope
OR
eyepiece lens ✔
Using a converging lens as a magnifying glass was the most common correct answer.
The student mounts the same lens on a ruler and light from a distant object is incident on the lens.
Identify, with a vertical line, the position of the focussed image. Label the position I.
[1]
I labelled at 25 cm mark ✔
Another very easy and well answered ray diagram question.
The image at I is the object for a second converging lens. This second lens forms a final image at infinity with an overall angular magnification for the two lens arrangement of 5. Calculate the distance between the two converging lenses.
[2]
the second lens has «cm» ✔
«so for telescope image to be at infinity»
the second lens is placed at 27 «cm»
OR
separation becomes 12 «cm» ✔
Only candidates who realised that a simple telescope was being constructed were able to answer the question correctly. Most candidates realised that the focal lenses need to be added but few found the focal lens of the second lens correctly.
A new object is placed a few meters to the left of the original lens. The student adjusts spacing of the lenses to form a virtual image at infinity of the new object. Outline, without calculation, the required change to the lens separation.
[2]
image formed by 10 cm lens is greater than 10 cm/further to the right of the first lens ✔
so second lens must also move to the right OR lens separation increases ✔
Award [1 max] for bald “separation increases”.
Many candidates did not read the question carefully and provided totally incorrect answers. It does not seem to be generally well known that if a distant object is moved to the right, for a converging lens, then the real image must also move to the right.