Question

In a Young's Double Slit Experiment, if the distance between the slits is halved and the distance to the screen is doubled, what happens to the fringe width?

• A. Doubled
• B. Halved
• C. Quadrupled
• D. Unchanged

Hint:

Fringe width depends on both the screen distance and the slit separation together, do not work out the effect of each change separately and then add them, combine the two ratios by dividing one by the other. Also note that here the slit separation is being reduced, not increased, which pushes the fringe width up even further.

Answer 1

The Correct Option is C

Concept: The fringe width in Young's Double Slit Experiment is given by \[ \beta = \frac{\lambda D}{d} \] where
• \( \lambda \) = wavelength of light
• \( D \) = distance between slit and screen
• \( d \) = distance between the slits

Step 1: Write the original fringe width. \[ \beta = \frac{\lambda D}{d} \]

Step 2: Apply the new conditions. Distance to screen is doubled: \[ D' = 2D \] Distance between slits is halved: \[ d' = \frac{d}{2} \]

Step 3: Substitute into the fringe width formula. \[ \beta' = \frac{\lambda D'}{d'} \] \[ \beta' = \frac{\lambda (2D)}{d/2} \] \[ \beta' = \frac{2\lambda D}{d/2} \] \[ \beta' = 4 \frac{\lambda D}{d} \] \[ \beta' = 4\beta \] Thus, the fringe width becomes four times the original value. \[ \boxed{\beta' = 4\beta} \]

Answer 2

Concept:

• Fringe width in YDSE is directly proportional to the screen distance $D$ and inversely proportional to the slit separation $d$, since $\beta = \dfrac{\lambda D}{d}$ and the wavelength $\lambda$ stays fixed for a given source.
• When both $D$ and $d$ change together, the new fringe width can be found quickly using the ratio $\dfrac{\beta_{new}}{\beta} = \dfrac{D_{new}/D}{d_{new}/d}$, without re-deriving the whole formula.

Step 1: Write the ratio of new to old fringe width using proportionality.
$\beta \propto \dfrac{D}{d}$ so $\dfrac{\beta_{new}}{\beta} = \dfrac{D_{new}/D}{d_{new}/d}$

Step 2: Substitute the given changes into the ratio.
$D_{new} = 2D \Rightarrow \dfrac{D_{new}}{D} = 2$
$d_{new} = \dfrac{d}{2} \Rightarrow \dfrac{d_{new}}{d} = \dfrac{1}{2}$

Step 3: Compute the ratio to find the new fringe width.
$\dfrac{\beta_{new}}{\beta} = \dfrac{2}{1/2} = 4$
$\beta_{new} = 4\beta$

Final Answer: The fringe width becomes four times (quadrupled) the original value.