Question

A thin transparent sheet is placed in front of one slit of a Young's double slit. The fringe width will:

• A. increase
• B. decrease
• C. remain same
• D. none of these

Hint:

The sheet adds a constant extra path \((\mu-1)t\), which only shifts the pattern. Fringe width \(\lambda D/d\) has no sheet term.

Answer 1

The Correct Option is C

Step 1: Concept. In Young's double-slit experiment, the fringe width is \[\beta = \frac{\lambda D}{d},\] where \(\lambda\) is the wavelength of light, \(D\) is the slit-to-screen distance and \(d\) is the separation between the two slits.
Step 2: Effect of the thin sheet. Placing a thin transparent sheet in front of one slit introduces an extra optical path \((\mu - 1)t\) for that slit only. This adds a CONSTANT extra path difference for all points on the screen.
Step 3: Consequence. A constant added path difference shifts the ENTIRE fringe pattern sideways (the central bright fringe moves toward the covered slit), but it does not change \(\lambda\), \(D\) or \(d\).
Step 4: Since \(\beta = \lambda D/d\) contains none of the sheet parameters, the spacing between consecutive fringes is unchanged. The fringe width remains the same, option (iii).
Why other options are wrong: The sheet only translates the pattern; it neither compresses nor expands the fringe spacing, so 'increase' and 'decrease' are wrong.
\[\boxed{\beta = \frac{\lambda D}{d}\ \text{unchanged}}\]