Fig. 1 shows a student using a converging lens to project a sharp image of an illuminated arrow onto a screen inside a dark box. The object is 18 cm from th...
Question 1 Report
Fig. 1 shows a student using a converging lens to project a sharp image of an illuminated arrow onto a screen inside a dark box. The object is 18 cm from the lens. The screen is 36 cm from the lens.
(a) State whether the image on the screen is real or virtual. [1]
(b) Calculate the magnification of the image. [2]
(c) Describe the orientation of the image compared with the arrow. [1]
(d) Explain why the screen must be moved if the student moves the object closer to the lens. [3]
(e) Draw two rays on Fig. 1 to show how the top of the object forms the image. [4]
Answer Details
(a) The image on a screen is real, because rays actually meet there. [1]
(b) \[\text{magnification}=\frac{\text{image distance}}{\text{object distance}}=\frac{36\text{ cm}}{18\text{ cm}}=2\] [2]
(c) The image is inverted compared with the arrow. [1]
(d) Moving the object changes the image distance from the lens. The refracted rays then meet at a different position, so the screen must be moved to this new image position to give a sharp image. [3]
(e) The ray diagram needs a ray from the top of the arrow parallel to the principal axis, refracted through the far focal point, and a ray through the centre of the lens that continues undeviated. Their intersection on the screen gives the top of the image. [4]
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