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(B) The angular width of the central maximum in single slit diffraction is given by $	heta = \frac{2\lambda}{a}$, where $a$ is the slit width.Thus, $

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$4200$

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(B) The angular width of the central maximum in single slit diffraction is given by $	heta = \frac{2\lambda}{a}$, where $a$ is the slit width.Thus, $	heta \propto \lambda$.Let the initial wavelength be $\lambda_1 = 6000 	ext{ Å}$ and the initial angular width be $	heta_1 = 	heta$.Then $	heta_1 = k \lambda_1$ $(i)$When the wavelength is changed to $\lambda_2 = \lambda$, the angular width decreases by $30 \%$.So, $	heta_2 = 	heta_1 - 0.30 	heta_1 = 0.70 	heta_1$.Since $	heta_2 = k \lambda_2$, we have $0.70 	heta_1 = k \lambda_2$ (ii)Dividing equation (ii) by equation $(i)$:$\frac{k \lambda_2}{k \lambda_1} = \frac{0.70 	heta_1}{	heta_1}$$\frac{\lambda_2}{6000 	ext{ Å}} = 0.70$$\lambda_2 = 0.70 	imes 6000 	ext{ Å} = 4200 	ext{ Å}$.

In a single slit diffraction, the slit is illuminated with light of wavelength $6000 \text{ Å}$. If the slit were illuminated by a light of wavelength $\lambda$ the angular width of the central maximum decreases by $30 \%$. Then $\lambda$ is (in $\text{ Å}$)

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