Q 12-10-151JEE MainJEE Main 2017 (8 Apr)Easy
A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $\lambda$. If the second and fourth minima in the diffraction pattern at a distance $1$ m from the slit are at $3$ cm and $6$ cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)
Answer: (D) $3.0$ cm
Minima are at $y_n = \dfrac{n\lambda D}{b}$, equally spaced by $\dfrac{\lambda D}{b}$:
$$y_4 - y_2 = 2\frac{\lambda D}{b} = 3\ \text{cm} \;\Rightarrow\; \frac{\lambda D}{b} = 1.5\ \text{cm}$$
Central maximum width $= 2\dfrac{\lambda D}{b} = 3.0$ cm.
Solution by Sreeraj P, M.Sc Physics