Questions
Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern,
detailed solution
Correct option is C
As, intensity I∝ width of slit W Also, intensity I∝ square of amplitude A∴ I1I2=W1W2=A12A22 But W1W2=125 (given) ∴A12A22=125 or A1A2=125=15∴ImaxImin=A1+A22A1-A22=A1A2+12A1A2-12=15+1215-12=652-452=3616=94Similar Questions
In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness made of material of refractive index 1.2. Then intensity of light at the mid point of the source is
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