In this experiment, you will draw a ray diagram for a converging lens. You draws the lens with a focal length of 3.0 cm on a sheet of graph paper. You place...
Question 1 Report
In this experiment, you will draw a ray diagram for a converging lens. You draws the lens with a focal length of 3.0 cm on a sheet of graph paper. You places an object arrow of height 2.0 cm at a distance of 5.0 cm from the lens. The partially completed ray diagram is shown in Fig. 5.1.
The scale is 1 cm on the diagram = 1 cm actual size. F marks the focal points and 2F marks twice the focal length from the lens centre C.
(a) Record the object distance u from the diagram. [1]
(b) On Fig. 5.1, draw the second construction ray from the top of the object through the centre of the lens. [1]
(c) On Fig. 5.1, draw the third construction ray from the top of the object through F on the near side of the lens, and then parallel to the principal axis after the lens. [1]
(d) Mark and label the position of the image where the rays cross. [1]
(e) Measure and record the image distance v from the diagram. [1]
(f) Measure and record the image height from the diagram. [1]
(g) Calculate the magnification using M = v / u. [1]
(h) Describe the image in terms of its size, orientation and type (real or virtual). [2]
(i) Verify that your values satisfy the equation 1/f = 1/u + 1/v. Show your working. [2]
Answer Details
(a) The object distance measured from the diagram is:
\[ u = 5.0 \text{ cm} \] [1]
(The object is at 5.0 cm from the lens centre C, which lies between F (3.0 cm) and 2F (6.0 cm).)
(b), (c), (d) The completed ray diagram showing all three construction rays and the image:
Ray 1 (already partially drawn): travels parallel to the principal axis from the top of the object to the lens, then refracts to pass through F on the far side. [given]
Ray 2: travels from the top of the object straight through the centre C of the lens and continues in the same direction (undeviated). [1]
Ray 3: travels from the top of the object through F on the near side of the lens, then refracts to emerge parallel to the principal axis. [1]
The three rays converge at a point below the principal axis. The image is marked where the rays cross. [1]
(e) The image distance measured from the diagram:
\[ v = 7.5 \text{ cm} \] [1]
(Accept 7.0 to 8.0 cm.)
(f) The image height measured from the diagram:
\[ \text{image height} = 3.0 \text{ cm} \] [1]
(Accept 2.8 to 3.2 cm.)
(g) The magnification:
\[ M = \frac{v}{u} = \frac{7.5}{5.0} = 1.5 \] [1]
(Accept value consistent with measured \( v \).)
This means the image is 1.5 times the size of the object, consistent with the measured image height of 3.0 cm from a 2.0 cm object.
(h) The image is: [2]
- Enlarged (magnification > 1; the image is larger than the object). [1]
- Inverted (upside down) and real (it can be projected onto a screen, and the rays actually converge at the image position). [1]
The object is between F and 2F, which always produces an enlarged, inverted, real image beyond 2F on the other side of the lens.
(i) Verifying the thin lens equation \( 1/f = 1/u + 1/v \):
\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{5.0} + \frac{1}{7.5} = 0.200 + 0.133 = 0.333 \] [1]
\[ \frac{1}{f} = \frac{1}{3.0} = 0.333 \]
The two values agree (both equal 0.333), which verifies the lens equation. [1]
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