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- Wave Optics
NEET · Physics · Wave Optics
In a Young's double-slit experiment setup (Case 1), two slits are...
In a Young's double-slit experiment setup (Case 1), two slits are illuminated by a single monochromatic source, acting as coherent sources. The maximum intensity observed on the screen is I_\max. In another setup (Case 2), the two slits are replaced by two independent, incoherent monochromatic sources, each providing the same individual intensity on the screen as the slits did in Case 1. The resulting uniform intensity on the screen is . What is the ratio of I_\max to ?
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let the individual intensity of each source be .
For Case 1 (coherent sources), the resultant intensity is given by .
The maximum intensity occurs when the phase difference .
\Rightarrow I_\max = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos(0) = 4I_0.
For Case 2 (incoherent sources), the phase difference varies randomly and rapidly with time. The average value of is zero.
Thus, the resultant uniform intensity is the simple sum of the individual intensities.
.
The ratio is \dfrac{I_\max}{I_{uni}} = \dfrac{4I_0}{2I_0} = \dfrac{2}{1}.
Answer:
For Case 1 (coherent sources), the resultant intensity is given by .
The maximum intensity occurs when the phase difference .
\Rightarrow I_\max = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos(0) = 4I_0.
For Case 2 (incoherent sources), the phase difference varies randomly and rapidly with time. The average value of is zero.
Thus, the resultant uniform intensity is the simple sum of the individual intensities.
.
The ratio is \dfrac{I_\max}{I_{uni}} = \dfrac{4I_0}{2I_0} = \dfrac{2}{1}.
Answer: