In a diffraction pattern due to a single slit | vedclass

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(C) For the first minimum in a single slit diffraction pattern,the condition is given by $a \sin 	heta = n \lambda$. For the first minimum,$n = 1$,so

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$sin^{-1} \left( \frac{3}{4} \right)$

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(C) For the first minimum in a single slit diffraction pattern,the condition is given by $a \sin 	heta = n \lambda$. For the first minimum,$n = 1$,so $a \sin 	heta = \lambda$.Given $	heta = 30^{\circ}$,we have $\sin 30^{\circ} = \frac{1}{2}$.Thus,$a (\frac{1}{2}) = \lambda$,which implies $a = 2 \lambda$ ..... $(i)$.For the first secondary maximum,the condition is $a \sin 	heta' = (n + \frac{1}{2}) \lambda$ where $n = 1$. So,$a \sin 	heta' = \frac{3}{2} \lambda$.Substituting $a = 2 \lambda$ from equation $(i)$ into this expression:$(2 \lambda) \sin 	heta' = \frac{3}{2} \lambda$.Dividing both sides by $2 \lambda$,we get $\sin 	heta' = \frac{3}{4}$.Therefore,$	heta' = \sin^{-1} \left( \frac{3}{4} ight)$.

In a diffraction pattern due to a single slit of width $a$,the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \; \mathring{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

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