Physics - Wave Optics Question with Solution | TestHub

Fringe Width: $\beta =\frac{\lambda D}{d}$

Path Difference ($\Delta x$): $\Delta x=\frac{d\cdot y}{D}$

Central Maxima Intensity: $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

(P) Source $S$ moves up

$\Delta x_{total}=\Delta x_{source}+\Delta x_{slits}$

Central fringe shifts downward to keep $\Delta x=0$. $\beta$ is unchanged. (2), (3)

(Q) Glass slab at $S_1$

$\Delta x=(\mu -1)t$

Central fringe shifts upward towards the slab. $\beta$ is unchanged. (1), (3)

(R) Immersed in water

${\lambda }^{\prime }=\frac{\lambda }{{\mu }_w}$

Since $\beta \propto \lambda$, the fringe width decreases. (4)

(S) $S_2$ diameter $\uparrow$

$I\propto width⟹I_2\uparrow$

$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$ increases. (5)