Question

In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength \(5000\) Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength \(6400\) Å. Now, to reduce the fringe width by \(20\%\) of its initial value, the screen should be moved

• A. towards the source by \(37.5\) cm
• B. towards the source by \(62.5\) cm
• C. away from the source by \(62.5\) cm
• D. away from the source by \(37.5\) cm

Hint:

Fringe width is proportional to lambda times D, so set the new product to 80 percent of the old one.

Answer 1

The Correct Option is A

Step 1: Ratio method
$\dfrac{\beta_2}{\beta_1} = \dfrac{\lambda_2D_2}{\lambda_1D_1} = 0.8$.

Step 2: Substitute
$\dfrac{6400}{5000}\cdot\dfrac{D_2}{100} = 0.8$, so $1.28\times\dfrac{D_2}{100} = 0.8$.

Step 3: Solve
$D_2 = \dfrac{80}{1.28} = 62.5$ cm.

Step 4: Direction
$D$ falls from 100 cm to 62.5 cm, so the screen shifts closer to the source by $100 - 62.5 = 37.5$ cm.

Final Answer:
The screen moves 37.5 cm towards the source. This is option (A). \[ \boxed{\text{(A) }\text{towards the source by 37.5 cm}} \]