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(A) Let the intensities at the slits be $I_1 = 4I_0$ and $I_2 = I_0$. The resultant intensity is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi = 5I_0 + 4

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$(A)$ and $(B)$

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(A) Let the intensities at the slits be $I_1 = 4I_0$ and $I_2 = I_0$. The resultant intensity is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi = 5I_0 + 4I_0 \cos \phi$.For maxima, $\cos \phi = 1$, so $I_{max} = 9I_0$. For minima, $\cos \phi = -1$, so $I_{min} = I_0$.Condition for maxima is $d \sin 	heta = n\lambda$, where $n = 0, \pm 1, \pm 2, \dots$.$(A)$ If $d = \lambda$, then $\sin 	heta = n$. For $n = 0$, $	heta = 0$ (central maximum). For $n = \pm 1$, $\sin 	heta = \pm 1$, so $	heta = \pm 90^\circ$. These are at infinity, so only the central maximum is observed on the screen. Thus, $(A)$ is correct.$(B)$ If $\lambda $(C)$ If $I_1$ is reduced to $I_0$, then $I_1 = I_2 = I_0$. $I_{max} = 4I_0$ (decreases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). Thus, $(C)$ is incorrect.$(D)$ If $I_2$ is increased to $4I_0$, then $I_1 = I_2 = 4I_0$. $I_{max} = 16I_0$ (increases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). The bright fringes increase, but dark fringes decrease. Thus, $(D)$ is incorrect.

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