

Now net electric field between plates
$\overrightarrow{\mathrm{E}}_{\mathrm{net}} =\mathrm{E} \cos 60^{\circ}(-\hat{\mathrm{x}})+\left(\mathrm{E}-\mathrm{E} \sin 60^{\circ}\right)(\hat{\mathrm{y}})$
$=\frac{\sigma}{2 \varepsilon_{0}}\left[-\frac{\hat{\mathrm{x}}}{2}+\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}\right]$
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$(A)$ If $\mathrm{d}=\lambda$, the screen will contain only one maximum
$(B)$ If $\lambda<\mathrm{d} < 2 \lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
$(C)$ If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$ , the intensities of the observed dark and bright fringes will increase
$(D)$ If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$ , the intensities of the observed dark and bright fringes will increase