Here are two questions from wave optics. The first question is meant for testing your understanding of coherent sources and interference while the second one is meant for testing your knowledge of the interference pattern produced by Young’s double slit.
(1) Two coherent sources emitting light of wave length λ and amplitude A produce an interference pattern on a screen. The maximum intensity of the interference pattern is I. If the light sources are not coherent but the wave length and amplitude are λ and A respectively, the maximum intensity on the screen will be
(a) I/4
(b) I/2
(c) I
(d) √I
(e) 2√I
At the interference maximum the amplitudes of the light waves get added. Therefore, the resultant amplitude at the interference maximum is A + A = 2A.
The intensity is directly proportional to the
I α 4A2
Or, I = k×4A2……………(i)
where k is the constant of proportionality
Since the amplitude of each light wave is A, the intensity (I0) of light produced by each source is given by
I0 = k×A2
This gives I0 = I/4 from (i).
Each source will produce uniform intensity I0 everywhere on the screen and the resultant intensity everywhere will be 2I0 (since there are two sources) and we have
2I0 = 2×I/4 = I/2
The correct option is (b).
(2) The angular width of the interference fringes obtained in a double slit experiment using light of wave length λ is found to be θ. If the entire experimental arrangement is immersed in water having refractive
(a) θ/4
(b) θ/3
(c) θ
(d) 4θ/3
(e) 3θ/4
The angular fringe width is λ/d where d is the separation between the slits. Therefore we have
θ = λ/d.
When the arrangement is submerged in water of refractive index n, the wave length of light is reduced to λ' given by
λ'= λ/n = 3λ/4 since n = 4/3
The angular fringe width now becomes θ'= λ'/d = 3λ/4d = 3θ/4.
You will find some useful multiple choice questions in this section at AP Physics Resources: Multiple Choice Questions on Interference and Diffraction .
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