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In the YDSE arrangement shown here, the intensity on the screen due to slit -2 is four times that of slit -1. If resultant intensity at the position of central maxima O is I, the resultant intensity at point P, where the phase difference between two waves coming from two slits is `cos^(-1)((1)/(4))` is

A

`(I)/(3)`

B

`(2I)/(3)`

C

`(I)/(4)`

D

`(I)/(2)`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the resultant intensity at point P in a Young's Double Slit Experiment (YDSE) setup, given that the intensity from slit 2 is four times that from slit 1. ### Step-by-Step Solution: 1. **Define Intensities from Each Slit:** Let the intensity from slit 1 be \( I_1 = I_0 \) and the intensity from slit 2 be \( I_2 = 4I_0 \). 2. **Resultant Intensity at Central Maxima (O):** The resultant intensity at the central maxima (point O) is given as \( I \). Using the formula for resultant intensity in YDSE: \[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \] At the central maxima, the phase difference \( \phi = 0 \) (since the path difference is zero), so \( \cos \phi = 1 \): \[ I = I_0 + 4I_0 + 2\sqrt{I_0 \cdot 4I_0} \cdot 1 \] \[ I = 5I_0 + 2 \cdot 2I_0 = 5I_0 + 4I_0 = 9I_0 \] 3. **Express \( I_0 \) in Terms of \( I \):** From the above, we have: \[ I = 9I_0 \implies I_0 = \frac{I}{9} \] 4. **Phase Difference at Point P:** The phase difference at point P is given as \( \phi = \cos^{-1}\left(\frac{1}{4}\right) \). 5. **Resultant Intensity at Point P:** We will use the same formula for resultant intensity at point P: \[ I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \] Substituting \( I_1 = I_0 \), \( I_2 = 4I_0 \), and \( \cos \phi = \frac{1}{4} \): \[ I_P = I_0 + 4I_0 + 2\sqrt{I_0 \cdot 4I_0} \cdot \frac{1}{4} \] \[ I_P = I_0 + 4I_0 + 2 \cdot 2I_0 \cdot \frac{1}{4} \] \[ I_P = 5I_0 + I_0 = 6I_0 \] 6. **Substituting \( I_0 \) in Terms of \( I \):** Now substituting \( I_0 = \frac{I}{9} \): \[ I_P = 6 \cdot \frac{I}{9} = \frac{2}{3} I \] ### Final Result: The resultant intensity at point P is: \[ \boxed{\frac{2}{3} I} \]
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