Question

In the interference experiment using a biprism, the distance of the slits from the screen is increased by 25% and the separation between the slits is halved. If 'W' represents the original fringewidth, the new fringewidth is

• A. 2 W
• B. 2.5 W
• C. 4 W
• D. 1.5 W

Hint:

Fringe width satisfies the proportionality relationship $W \propto \frac{D}{d}$. If $D$ is multiplied by $1.25$ and $d$ is multiplied by $0.5$, the total scaling factor is simply $\frac{1.25}{0.5} = 2.5$. Multiplying the initial width by this combined factor yields $2.5W$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the modified fringe width in a biprism interference arrangement when the slit-to-screen distance undergoes a 25% increase and the interval separating the slits is cut exactly in half.

Step 2: Key Formula or Approach:
The structural fringe width $W$ produced during double-slit or biprism interference is given by the relation:
$$W = \frac{\lambda D}{d}$$ where $\lambda$ represents the source light wavelength, $D$ represents the normal distance extending to the screen, and $d$ is the gap distance separating the virtual coherent sources.

Step 3: Detailed Explanation:
Let the initial fringe width setup be defined by the baseline equation:
$$W = \frac{\lambda D}{d}$$ Now, let's establish the modified operational parameters according to the conditions stated in the problem:
1. The distance between the slits and the screen $D$ is increased by 25%:
$$D' = D + 0.25D = 1.25D = \frac{5}{4}D$$ 2. The separation between the slits $d$ is halved:
$$d' = \frac{d}{2}$$ Substitute these modified parameters into the fringe width equation to evaluate the new fringe width $W'$:
$$W' = \frac{\lambda D'}{d'} = \frac{\lambda (1.25D)}{\frac{d}{2}}$$ Move the fraction from the denominator to the numerator as a multiplier:
$$W' = 2 \times 1.25 \times \left(\frac{\lambda D}{d}\right)$$ $$W' = 2.5 \left(\frac{\lambda D}{d}\right)$$ Substituting the original value $W$ gives:
$$W' = 2.5W$$

Step 4: Final Answer:
The new fringe width value is 2.5 W, which corresponds perfectly to option (B).