Question

In a Young's double slit experiment, constructive interference is produced at a certain point P. The intensities of light at P due to the individual sources are 4 and 9 units. The resultant intensity at point P will be –

• A. 13 units
• B. 25 units
• C. \sqrt{97} units
• D. 5 units

Answer 1

Answer: (B)

25 units

Answer 2

I_P = I_max =$\left( \sqrt{I_{1}} + \sqrt{I_{2}} \right)^{2}$= $\left( \sqrt{4} + \sqrt{9} \right)^{2}$= 25 unit

I_P = I_max =$\left( \sqrt{I_{1}} + \sqrt{I_{2}} \right)^{2}$= $\left( \sqrt{4} + \sqrt{9} \right)^{2}$= 25 unit