Question

In Young's double-slit experiment, interference fringes are obtained on a screen placed at a distance of $75\text{ cm}$ from the slits. When the separation between the two narrow slits is doubled, the fringe width decreases. In order to maintain the initial fringe width, the screen should be moved through a distance of

• A. $150\text{ cm}$ away from the slits
• B. $75\text{ cm}$ towards the slits
• C. $75\text{ cm}$ away from the slits
• D. $150\text{ cm}$ towards the slits

Hint:

Look directly at the proportionality: $\beta \propto \frac{D}{d}$. If you double the value in the denominator ($d$), you must exactly double the value in the numerator ($D$) to keep the overall fraction unchanged. Doubling $75\text{ cm}$ gives $150\text{ cm}$, meaning you added exactly $75\text{ cm}$ to the distance!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem outlines a scenario using Young's Double Slit Experiment (YDSE). Initially, the screen sits at a distance $D_1 = 75\text{ cm}$.
When the slit separation distance ($mm$-scale parameter $d$) is scaled up by a factor of 2, the width of individual fringes drops. We need to find how much and in which direction to shift the screen position ($\Delta D$) to restore the fringe width to its original value.

Step 2: Key Formula or Approach:
The formula for the fringe width ($\beta$) in an interference pattern is given by: $$\beta = \frac{\lambda D}{d}$$ Where $\lambda$ is the source wavelength, $D$ is the slit-to-screen distance, and $d$ is the distance separating the two slits.
To keep the fringe width constant ($\beta_1 = \beta_2$) using the same light source ($\lambda_1 = \lambda_2$), the ratio of the structural dimensions must remain invariant: $$\frac{D_1}{d_1} = \frac{D_2}{d_2}$$

Step 3: Detailed Explanation:
Identify the structural shifts from the problem text: Initial distance: $D_1 = 75\text{ cm}$ Initial slit width: $d_1 = d$ Secondary slit width: $d_2 = 2d$
Set up the equality equation to find the required new screen distance $D_2$: $$\frac{75}{d} = \frac{D_2}{2d}$$ Cancel out the common variable $d$ from both denominators: $$75 = \frac{D_2}{2} \implies D_2 = 75 \times 2 = 150\text{ cm}$$ Now, compute the net displacement shift ($\Delta D$) of the screen: $$\Delta D = D_2 - D_1 = 150\text{ cm} - 75\text{ cm} = 75\text{ cm}$$ Since $D_2 > D_1$ ($150\text{ cm} > 75\text{ cm}$), the screen must be shifted further away from the slit plane.

Step 4: Final Answer:
The screen should be moved through a distance of $75\text{ cm}$ away from the slits, matching option (C).