Question

The 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 $36\%$ by reducing its width then the intensity of light at the same point will be

• A. $90$ units
• B. $89$ units
• C. $67$ units
• D. $81$ units

Answer 1

The correct option is D $81$ units

Since , $I_{max}=4I_0$
Given, $I_{max}=100$
$\therefore$ Intensity of each source , $I_0=\frac{100}{4}=25\text{unit}$
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
$I=I_1+I_2+2\sqrt{I_1{I}_2}=25+16+2\sqrt{25\times 16}=81\text{unit}$