Fig. 1 shows a safety check being carried out in a theatre. A technician places a flat mirror on the floor so that a camera can view a cable tray above the ...
Question 1 Report
Fig. 1 shows a safety check being carried out in a theatre. A technician places a flat mirror on the floor so that a camera can view a cable tray above the stage. The cable tray is 1.6 m in front of the mirror. The camera is beside the tray, on the same side of the mirror. Light from the tray reflects from the mirror into the camera. The mirror does not produce sound, but the reflected light allows the technician to check the tray before electrical current is switched on.
(a) What is meant by the term virtual image? [2]
(b) Describe two properties of the image of the cable tray, other than that it is virtual. [2]
(c) Calculate the distance between the cable tray and its image. Give your answer in m. [2]
(d) Draw on Fig. 1 one further ray from the cable tray that reflects from the mirror and enters the camera. [2]
(e) Explain why the image cannot be formed on a screen placed behind the mirror. [2]
Answer Details
(a) A virtual image is formed at a position where light rays appear to come from. [1] The rays do not actually pass through or meet at that image position behind the mirror. [1]
(b) Two properties are that the image is upright [1] and the same size as the cable tray. [1] It is also laterally inverted and is the same distance behind the mirror as the object is in front.
(c) The cable tray is 1.6 m in front of the mirror, so its image is 1.6 m behind the mirror. [1]
\[1.6\text{ m}+1.6\text{ m}=3.2\text{ m}\]
The distance between the cable tray and its image is 3.2 m. [1]
(d) A further ray must travel from the cable tray to the mirror and then reflect into the camera. [1] It must obey \(i=r\). [1]
(e) The reflected rays diverge and do not actually meet behind the mirror. [1] Therefore no light falls on a screen at the apparent image position, so the image cannot be formed on that screen. [1]
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