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Q. Two coherent sources of equal intensity produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $36 \%$ reducing its width, then the intensity of light at the same point will be

Wave Optics

Solution:

Intensity of each source $=I_{0}=\frac{100}{4}=25$ unit
If the intensity of one of the source is reduced by $36 \%$ then $I_{1}=25$ unit and
$I_{2}=25-25 \times \frac{36}{100}=16($ unit $)$
Hence resultant intensity at the same point will be
$I=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$
$=25+16+2 \sqrt{25 \times 16}=81$ unit