Correct Option: (i) will decrease, but the intensity of bright fringe remains the same.
Step 1 (Formula / Governing Law & Given Data): [1 Mark]
The fringe width ($\beta$) in Young's Double Slit Experiment (YDSE) is given by the formula:
$$\beta = \frac{\lambda D}{d}$$Where:
- $\lambda$ = Wavelength of the monochromatic light used.
- $D$ = Distance between the slits and the screen.
- $d$ = Distance between the two coherent slits $S_1$ and $S_2$.
Step 2 (Step-by-Step Mathematical Working): [1 Mark]
According to the problem, the screen is moved closer to the plane of the slits. This implies that the distance $D$ is decreased ($D' < D$).
Since $\lambda$ and $d$ remain constant, we analyze the relationship:
$$\beta \propto D$$As $D$ decreases, the fringe width $\beta$ must also decrease proportionally. Regarding intensity, the intensity of the bright fringe ($I_{max}$) is given by $I_{max} = 4I_0$ (where $I_0$ is the intensity of individual slits). Since the amplitude of the waves reaching the screen does not change significantly with a small shift in distance $D$, the intensity of the bright fringes remains constant.
Step 3 (Final Concluded Answer): [1 Mark]
Since $\beta = \frac{\lambda D}{d}$, a decrease in $D$ leads to a decrease in $\beta$. The intensity of the bright fringes is independent of the screen distance $D$ in this experimental configuration. Therefore, the fringe width decreases while the intensity remains the same.
💡 EL10 Concept Insight:
Think of the interference pattern like a projection on a wall; if you move the wall closer to the projector, the image gets smaller (fringe width decreases), but the brightness of the light hitting the wall stays just as intense as it was before.