The maximum number of possible interference maxima when slit separation is equal to 4 times the wavelength of light used in a double slit experiment is

In double slit experiment, for possible interference maxima on the screen is given as $$ \begin{aligned} d \sin \theta &=n \lambda \\ \text { Given, slit width, } d &=4 \lambda \\ \therefore \quad 4 \lambda \sin \theta &=n \lambda \\ \Rightarrow \quad 4 \sin \theta &=n \end{aligned} $$ Since, maximum value of $\sin \theta=1$ $\therefore$ For maximum number of possible interference maxima, $$ \begin{aligned} & & 4 \cdot 1 &=n \\ \Rightarrow & & n &=4 \\ \text { i.e., } & & n &=0,1,2,3 \end{aligned} $$ $\therefore$ Maximum number of possible maxima will be 7 i.e. $0, \pm 1, \pm 2, \pm 3$ Hence, no option is correct.