Question

In a biprism experiment, monochromatic light of wavelength $\lambda$ is used. The distance between two coherent sources is kept constant. If the distance between slit and eyepiece is varied as $D_1, D_2, D_3$ and $D_4$, the corresponding measured fringe widths are $z_1, z_2, z_3$ and $z_4$, then

• A. $\dfrac{z_1}{D_1} = \dfrac{z_2}{D_2} = \dfrac{z_3}{D_3} = \dfrac{z_4}{D_4}$
• B. $z_1 D_1 = z_2 D_2 = z_3 D_3 = z_4 D_4$
• C. $z_1 \sqrt{D_1} = z_2 \sqrt{D_2} = z_3 \sqrt{D_3} = z_4 \sqrt{D_4}$
• D. $z_1 D_1^2 = z_2 D_2^2 = z_3 D_3^2 = z_4 D_4^2$

Hint:

Fringe width in interference experiments is directly proportional to the screen distance.

Answer 1

The Correct Option is A

Step 1: Expression for fringe width.
In Young’s double slit (biprism) experiment, fringe width is given by:
\[ \beta = \frac{\lambda D}{d} \] where $D$ is the distance between slit and screen and $d$ is the separation of coherent sources.

Step 2: Analyze given conditions.
Here, $\lambda$ and $d$ are constant, while $D$ varies.
Hence,
\[ \beta \propto D \]

Step 3: Write proportionality relation.
\[ \frac{z}{D} = \text{constant} \]

Step 4: Conclusion.
\[ \frac{z_1}{D_1} = \frac{z_2}{D_2} = \frac{z_3}{D_3} = \frac{z_4}{D_4} \]