Question

Figure shows the plane (top view) of a cubical room, with wall CD as a plane mirror; each side of the room is 3metres in length. A camera P is placed at the midpoint of the wall AB. At what distance should the camera be focused to photograph the image of an object placed at A?

Answer 1

Hint: When light falls on mirror then it reflected back in the same medium .So when light falls from camera then it reflected back and image of that point will be formed through plane mirror and Plane mirror always forms the virtual image of an object at distance equal to distance of object from mirror.

Complete answer:

Each side of the square is\[3cm\].
Camera P is placed at midpoint of wall AB such that \[AP=1.5cm\]
Since Mirror is placed at CD so light coming from the camera reflects at point o and image of point A is formed at A’ such that\[AD=D{{A}^{'}}=3cm\].
Let us assume that the camera should be focused to photograph the image of an object A at\[x\].
In right angled triangle APA’
\[{{({{A}^{'}}P)}^{2}}={{(AP)}^{2}}+{{(A{{A}^{'}})}^{2}}\]
\[{{x}^{2}}={{(1.5)}^{2}}+{{(6)}^{2}}\]
\[{{x}^{2}}=2.25+36\]
\[{{x}^{2}}=38.25\]
\[x=6.18m\]
So the camera should be focused at distance \[6.18m\] to photograph the image of the object at A.

Note:
In the phenomenon of reflection of light- velocity, frequency & wavelength of light does not change even after the reflection and angle of incidence is always equal to angle of reflection. When two rays appear to meet at a point then after reflection image is formed of object A at point A’ then virtual image is formed. So we will calculate distance from P to A’ rather than from A to A’ because we have to calculate distance of image focused by camera rather than distance of object from image.