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(A) The angular width of the central maximum in single slit diffraction is given by $	heta = \frac{2\lambda}{a}$, and the linear width is $\beta_{cm}

Vedclass · Accepted Answer

$A$, $C$, $E$ only

Vedclass · Answer

(A) The angular width of the central maximum in single slit diffraction is given by $	heta = \frac{2\lambda}{a}$, and the linear width is $\beta_{cm} = \frac{2\lambda D}{a}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $a$ is the slit width.$(A)$ Since $\beta_{cm} \propto \lambda$, the width of the central maximum increases as the wavelength increases. Thus, $(A)$ is correct.$(B)$ Since $\beta_{cm} \propto \lambda$, the width decreases as wavelength decreases. Thus, $(B)$ is incorrect.$(C)$ Since $\beta_{cm} \propto \frac{1}{a}$, the width increases as the slit width $a$ decreases. Thus, $(C)$ is correct.$(D)$ Since $\beta_{cm} \propto \frac{1}{a}$, the width decreases as the slit width $a$ increases. Thus, $(D)$ is incorrect.$(E)$ The intensity of the central maximum is proportional to the square of the slit width and inversely proportional to the square of the wavelength $(	ext{Intensity} \propto \frac{a^2}{\lambda^2})$. Therefore, as $\lambda$ decreases, the intensity (brightness) of the central maximum increases. Thus, $(E)$ is correct.Conclusion: Statements $(A)$, $(C)$, and $(E)$ are correct.

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