Wave Optics: NEET previous year questions

The interference intensity follows $I = I_0\cos^2\left(\dfrac{\Delta\phi}{2}\right)$. A path difference of $\lambda$ gives $\Delta\phi = 2\pi$, so $I = I_0 = K$, i.e. $I_0 = K$. For a path difference $\lambda/3$, $\Delta\phi = \dfrac{2\pi}{3}$, giving $I = I_0\cos^2\left(\dfrac{\pi}{3}\right) = \dfrac{I_0}{4} = \dfrac{K}{4}$ (option (1)). The other choices ignore the $\cos^2$ dependence on phase.

Interference and diffraction only redistribute light energy, so darkening of one region is compensated by brightening of another, keeping total energy unchanged. This makes statement I consistent with conservation of energy, i.e. Statement I is true. However, interference and diffraction are general wave phenomena shown by sound, water and other waves, not only light, so statement II is false. Hence 'I true, II false' (option (3)).

The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the separation between the slits. If $d$ is halved and $D$ is doubled, $\beta$ becomes $\beta' = \frac{\lambda (2D)}{\frac{d}{2}} = 4 \beta$. Therefore, the fringe width becomes four times. Chapter: Wave Optics.