Question:easy

In Young's double-slit experiment, if the separation between the slits is halved and the distance between the slits and the screen is doubled, the fringe width will be:

Show Hint

Fringe width is directly proportional to \(D\) and inversely proportional to \(d\). Any fractional changes can be directly multiplied: \((2) \times (1 / 0.5) = 4\).
Updated On: Jun 15, 2026
  • Halved
  • Unchanged
  • Doubled
  • Quadrupled
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understand the change being made.
In a double-slit setup the slit gap is cut to half, while the screen is moved twice as far away. We want the effect on the fringe width.
Step 2: Recall the fringe width formula.
The spacing between bright fringes is \[ \beta = \frac{\lambda D}{d} \] where $\lambda$ is the wavelength, $D$ the slit to screen distance, and $d$ the slit separation.
Step 3: Note that $\lambda$ is fixed.
The same light source is used, so the wavelength does not change. Only $D$ and $d$ are being altered.
Step 4: Apply the new values.
The new distance is $D' = 2D$ and the new separation is $d' = d/2$. Substitute: \[ \beta' = \frac{\lambda (2D)}{(d/2)} \]
Step 5: Simplify the fraction.
Dividing by $d/2$ is the same as multiplying by $2/d$, so \[ \beta' = \frac{2\lambda D \cdot 2}{d} = 4\,\frac{\lambda D}{d} = 4\beta \]
Step 6: Interpret the result.
The fringe width grows to four times its original value, so it is quadrupled, which is option (D).
\[ \boxed{\beta' = 4\beta} \]
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