1) You look through a camera toward an image of a hummingbird in a plane mirror.

Question

1) You look through a camera toward an image of a hummingbird in a plane mirror. The camera is 4.89 m in front of the mirror. The bird is at camera level, 5.95 m to your right and 4.26 m from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

2) In the figure, an isotropic point source of light S is positioned at distance d from a viewing screen A and the light intensity IP at point P (level with S) is measured. Then a plane mirror M is placed behind S at distance 4.2d. By how much is IP multiplied by the presence of the mirror?

Explanation / Answer

1)

Distance from the camera to the mirror, d1 = 4.89 m

Distance of the bird infront of the mirror, d2 = 4.26 m

Lateral distance between camera and the bird, d3 = 5.95 m

Solution:

A right angled triangle can be constructed with

sides as d3 and distance between the camera and

image plane (d1 + d2 ).

Then, the diagonal of the triangle gives the focus

distance.

Focus distance, d = ? [ ( d1 + d2)^2 + d3^2 ]

                              = ? [ (4.89 + 4.26)^2 + 5.95^ 2 ]

                              = 10.91 m

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