In Young's double slit experiment, the width | vedclass

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(A) Let the width of the first slit be $w_1 = w$ and the second slit be $w_2 = 2w$. Since intensity $I \propto 	ext{width}$, we have $I_1 = I_0$ and

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$\frac{I_m}{9}(1 + 8 \cos^2 \frac{\phi}{2})$

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(A) Let the width of the first slit be $w_1 = w$ and the second slit be $w_2 = 2w$. Since intensity $I \propto 	ext{width}$, we have $I_1 = I_0$ and $I_2 = 2I_0$. Since $I \propto A^2$, the amplitudes are $A_1 = a$ and $A_2 = a\sqrt{2}$. The resultant intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi = I_0 + 2I_0 + 2\sqrt{2I_0^2} \cos \phi = 3I_0 + 2\sqrt{2} I_0 \cos \phi$. The maximum intensity $I_m$ occurs when $\cos \phi = 1$, so $I_m = 3I_0 + 2\sqrt{2} I_0 = I_0(3 + 2\sqrt{2})$. Using $\cos \phi = 2 \cos^2 \frac{\phi}{2} - 1$, we have $I = 3I_0 + 2\sqrt{2} I_0 (2 \cos^2 \frac{\phi}{2} - 1) = I_0(3 - 2\sqrt{2} + 4\sqrt{2} \cos^2 \frac{\phi}{2})$. Since $I_0 = \frac{I_m}{3 + 2\sqrt{2}}$, we rationalize: $I_0 = I_m(3 - 2\sqrt{2})$. Substituting this, $I = I_m(3 - 2\sqrt{2})(3 - 2\sqrt{2} + 4\sqrt{2} \cos^2 \frac{\phi}{2}) = I_m(1 + 8 \cos^2 \frac{\phi}{2}) / 9$.

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