- A$3300$
- ✓$4400$
- C$5500$
- D$6600$
$a\sin \theta = \left( {2n + 1\frac{\lambda }{2}} \right)$
and for diffraction minima,
$a \sin \theta=n \lambda$
According to question,
$\left(2 \times 1+1 \frac{\lambda}{2}\right)=1 \times 6600$
$\left( {\because {\lambda _{\text{R}}} = 6600\,\,\mathop {\text{A}}\limits^o } \right)$
${\lambda = \frac{{6600 \times 2}}{3}}$
$\,\lambda = 4400\,\mathop {\text{A}}\limits^o $
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ If $I_1=I_2$, then B' cannot be equal to zero at the origin $(0,0,0)$
$(B)$ If $\mathrm{I}_1>0$ and $\mathrm{I}_2<0$, then $\mathrm{B}$ can be equal to zero at the origin $(0,0,0)$
$(C)$ If $\mathrm{I}_1<0$ and $\mathrm{I}_2>0$, then $\mathrm{B}$ can be equal to zero at the origin $(0,0,0)$
$(D)$ If $\mathrm{I}_1=\mathrm{I}_2$, then the $\mathrm{z}$-component of the magnetic field at the centre of the loop is $\left(-\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\right)$


(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )