Questions
- What is meant by a virtual image?
- The figure below shows an object O being viewed using two inclined mirrors M1 and M2.
Complete the diagram by sketching rays to show the position of the image as seen by the eye E - The figure below shows an object O placed in front of a plane mirror
On the same diagram, draw rays to locate the position of the image I as seen from the eye E. - The diagram shows a ray of light incident on a plane mirror at point O.
The mirror is rotated clockwise through an angle of 300 about an axis perpendicular to the paper. Determine the angle through which the reflected ray rotated. - A luminous point object took 3 s to move from P to Q in front of a pinhole camera as shown below.
What is speed in cm/s of the image on the screen? - The diagram shows the image of a watch face in a plane mirror
What is the time shown on the watch face? -
- Give two main reasons why concave mirrors are unsuitable as driving mirrors
- State one disadvantage of a convex mirror as a driving mirror
- Explain why a concave mirror is suitable for use as a make up mirror.
- In the space provided below, sketch a labeled diagram to show how a pinhole camera forms an image of a vertical object placed in front of the pinhole (3 marks)
- A building standing 100m from a pinhole camera produces on the screen of the camera an image 5 cm high 10 cm behind the pinhole. Determine the actual height of the building. (3 marks)
Answers
- - Image that cannot be formed on screen.
- Always on the opposite side of the object -
-
- Angle of rotation of reflected ray = 2(angle of rotation of mirrors)
= 2x 300
=600 - Measure P1Q1 in cm (i.e. length of image on the screen as shown below)
Divide this value by 3 seconds i.e. velocity =distance/time - 4:05 p.m
-
- - Key form real inverted images
- Highly magnified images which give a wrong perception of object distance.
- Small field of view. - Very small images, giving the illusion that the objects are far away.
- - Key form real inverted images
- Can from magnified, erected images.
Where
o= object
h= pin-hole
u- Object distance
v- Image distance-
u =100m
hi= 0.5cm
v=10cm
v/u = hi/h0
∴ ho = hi x u/v
= 5m x 100m/10cm
h0=50m
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