InterviewSolution
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InterviewSolution
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A ball swings back and forth in front of a concave mirror. The motion of the ball is described approximately by the equation `x= f cos (omega)t`, where f is the focal length of the mirror and x is measured along the axis of mirror. The origin is taken at the centre of curvature of the mirror. (a) Derive an expression for the distance from the mirror of the image of the swinging ball. (b) At what point does the ball appear to coincide with its image? (c) What will be the lateral magnifiaction of the image of the ball at time `t= T/2`, where T is time period of oscillation? |
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Answer» Correct Answer - A::B::C::D A(a) At` t=t`, ` u=-(2f+x)` `=-(2f+f cos (omega)t)` Using the mirror formula, `1/v+1/u=1/f` We have, `(1/v) -(1)/(2f+f cos (omega)t) =(-1)/(f)` `:. v=-((2+cos(omega)t)/(1+cos(omega)t))f` i.e distance of image from mirror at time t is `((2+cos(omega)t)/(1+cos(omega)t))f` (b) Ball coincides with its image at centre of curvature, i.e. at x=0 (c) At `t=T//2` or, `(omega)t=(pi), x=-f` i.e. `u=-f` or ball is at focus. so its image is at `(prop)` `m=(prop)`. |
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