In a single-slit diffraction pattern,the inte | vedclass

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(B) In a single-slit diffraction pattern,the intensity of the central maximum is proportional to the square of the slit width $(I \propto b^2)$.Howeve

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$4I_0$

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(B) In a single-slit diffraction pattern,the intensity of the central maximum is proportional to the square of the slit width $(I \propto b^2)$.However,the angular width of the central maximum is given by $\Delta 	heta = 2\lambda / b$.When the slit width $b$ is doubled to $2b$,the angular width becomes half $(\Delta 	heta' = \Delta 	heta / 2)$.Since the total energy passing through the slit is proportional to the slit width $(E \propto b)$,and the area of the central maximum on the screen is proportional to the angular width,the intensity $I$ is defined as $Energy / Area$.Thus,$I \propto b / (1/b) = b^2$. Wait,let's re-evaluate: The amplitude $A$ at the center is proportional to $b$. Therefore,the intensity $I_0 = A^2$ is proportional to $b^2$.If the slit width $b$ is doubled,the amplitude $A$ becomes $2A$,and the new intensity $I' = (2A)^2 = 4I_0$.

In a single-slit diffraction pattern,the intensity of the central maximum is $I_0$. If the slit width is doubled,what will be the intensity of the central maximum?

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