it is a screen at it is a point light sources o a ray incident on mirror perpendicularly and return back this mirror is some rotated through 22.5°this distance is 7 m then find the distance of image formed on screen from o
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Answer:Suppose the number of revolutions are n. The angle between 2 positions of the rotating mirror =
2
1
×27degrees.
Since the angle of rotation of mirror is half the angle through which the reflected ray rotates.
The time taken by the mirror in rotating through an angle θ is given by t=
2πn
θ
=
2×180×n
13.5degrees
sec. -- Eqn 1
This is also the time taken by the light to travel from original point to the fixed mirror and back, thus
t=
c
2d
=
3×10
8
2×22500
[d=22.5km=22500mandc=3×10
8
m/s] -- Eqn 2
From eqns 1 and 2,
2×180×n
13.5
=
3×10
8
2×22500
or n =
2×180×2×22500
13.5×3×10
8
=
16200000
40.5×10
8
=
162×10
5
40500×10
5
=250revolutions/s.
Hence, the number of revolutions are 250 revolutions/s.
Explanation:
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