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(B) The angular width of the central maximum in a single-slit diffraction pattern is given by $\beta = \frac{2\lambda}{d}$,where $\lambda$ is the wave

Vedclass · Accepted Answer

$4200$

Vedclass · Answer

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by $\beta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the slit width.Since $\beta \propto \lambda$,we have $\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$.Given $\lambda_1 = 6000 \ \mathring A$ and the angular width decreases by $30\%$,the new angular width $\beta_2 = \beta_1 - 0.30\beta_1 = 0.70\beta_1$.Substituting these values: $\frac{\beta_1}{0.70\beta_1} = \frac{6000}{\lambda_2}$.$\frac{1}{0.70} = \frac{6000}{\lambda_2} \Rightarrow \lambda_2 = 6000 	imes 0.70 = 4200 \ \mathring A$.

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