First slide

Young's double slit Experiment

Question

In Young's double slit experiment width of fringe produced on the screen is $β$ . Now one of the slits is covered with a sheet of glass $(μ = 1.5)$ of thickness $t = 8 λ$ where $λ$ is the wavelength of light used. Then distance of the new position of central bright fringe from the previous position of central bright fringe is

Moderate

A

$β$
B

2 $β$
C

4 $β$
D

6 $β$

Solution

Fringe width $β = \frac{λD}{d}$ where D is screen separation and d = slit separation. When one slit is covered, fringe width remains unchanged and the net path difference between rays at new position of central bright fringe should be zero.
$\begin{array}{l} ∴ (μ - 1) t - dsinθ = 0 \Rightarrow dsinθ = (μ - 1) t \Rightarrow dtanθ = (μ - 1) t \\ \Rightarrow d . \frac{y}{D} = (μ - 1) t \Rightarrow y = (μ - 1) \frac{t}{λ} . \frac{λD}{d} = (1 .5 - 1) \frac{8 λ}{λ} . β = 4 β \end{array}$

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