Question 23M.3.HL.TZ1.12
| Date | May 2023 | Marks available | [Maximum mark: 10] | Reference code | 23M.3.HL.TZ1.12 |
| Level | HL | Paper | 3 | Time zone | TZ1 |
| Command term | Calculate, Describe, Determine, Draw, State | Question number | 12 | Adapted from | N/A |
A converging lens is placed between an object and a screen. An image of the object is formed on the screen.
Draw a ray to locate the focal point of the lens. Label this point with the letter F.
[1]
one of the two rays above ✓
Converging lens. Ray optics, telescope, and spherical aberration proved to be well done, except for the detail in c) i).
The focal length of the lens is 4.0 cm and the height of the image is half the height of the object. Determine the distance of the object from the lens.
[3]
OR ✓
✓
u = 12 «cm» ✓
Diagram is not to scale so award [0] if answer obtained by measurement.
Allow MP1 if mistake in negative sign.
Do not allow ECF from MP1.
Award [3] for BCA.
The lens suffers from spherical aberration.
Draw lines to complete the rays in the diagram.
[1]
the extreme ray crosses principal axis closer than paraxial ray ✓
Describe the effect of spherical aberration on the image formed by the lens.
[1]
image is curved / blurred / distorted / poorly focused ✓
State how spherical aberration may be corrected.
[1]
block non-paraxial rays/ reduce aperture/use rays closer to axis/OWTTE
OR
use aspherical lens ✓
Allow parabolic lens.
Allow use of additional lens OR compensation plates.
A simple optical astronomical refracting telescope consists of an objective lens of focal length 75 cm and an eyepiece of focal length 4.0 cm. The telescope is used to view the Moon. The Moon subtends an angle of 0.51° at the unaided eye.
Calculate the angle subtended by the Moon when viewed through the telescope.
[2]
angular magnification is = 18.75 ✓
angle = «18.75 × 0.51» = 9.6° ✓
The telescope is now turned around so that the eye of the observer is behind the objective lens. State the change, if any, in the image of the Moon.
[1]
It would be «much» smaller ✓