[II] Shining a planar wave on a double-slit arrangement...

[II] Shining a planar wave on a double-slit arrangement (left) produces an interference pattern on a screen (as shown below). When one slit is covered with a glass plate (right), the phases of the two emerging waves will be different because the wavelength is shorter in glass than in air. If the phase difference is 90°, how is the interference pattern, shown left, altered? double slit interference pattern double slit planar wave planar wave glass slide A. The pattern vanishes. B. The bright spots lie closer together. C. The bright spots are farther apart. D. There are no changes. E. Bright and dark spots are interchanged. F. None of the above

Transcript

00:01 Hello students in this question we are given that shining a plain wave on a double slit arrangement produces an interference pattern as seen on the screen below.

00:12 So let us see the figure.

00:13 So over here this is the figure which is on the left.

00:17 So this type of figure comes in place and when one slit is covered with glass plate as shown in the right the phases of the two emerging waves will be different because the wavelength is shorter in glass than in air because the wavelength will be shorter in glass and it will pass less and the face difference will be altered so if the face difference is 90 degree how is the interference pattern shown left altered so in this question basically we are given the face difference which is phi here face difference phi is pi over to that is 90 degree now we know that here i resultant that is a intensity resultant here or we can write ir is equal to i1 plus i2 plus 2 i1 i2 cos 5 so similarly here we have at the intensity of the two waves are equal so we have ir is equal to i plus 2 i plus i both are equal plus 2 i squared cos y over 2 is 0 so we have i add is equal to 2 i this is our first equation now intensity will be minimum when 5 will be 0 for minimum intensity minimum intensity 5 should be equal to 0 because at 0 cost is cost function is maximum 5 should be be equal to 0 so cos 5 will be 1 so i max i max will be if we put cos 0 over here so it will be 2 it will be here it is basically over here root here it is also 2 i multiplied by i will be it will be over here i so i max will be 4i so let if we let i max equal to i0 so we have i not equal to 4 i so the intensity i will be 1 over 4th of the maximum intensity.

02:49 Now if we find the resultant intensity, so ir, which is equal to 2i, and i is i0 over 4, so 2 multiplied by i0 over 4.

03:02 So we can say ir is half of the maximum intensity...

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