Q129 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
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
- Correct answerAtowards the source by $37.5$ cm
- Btowards the source by $62.5$ cm
- Caway from the source by $62.5$ cm
- Daway from the source by $37.5$ cm
Solution
The initial fringe width is given by $\beta_1 = \dfrac{\lambda_1 D_1}{d}$
The final fringe width is given by $\beta_2 = \dfrac{\lambda_2 D_2}{d}$
It is given that the fringe width is reduced by $20\%$ of its initial value, so:
$\beta_2 = \beta_1 - 0.2\beta_1 = 0.8\beta_1$
Substituting the expressions for fringe widths:
$\dfrac{\lambda_2 D_2}{d} = 0.8 \dfrac{\lambda_1 D_1}{d}$
$D_2 = \dfrac{0.8 \lambda_1 D_1}{\lambda_2}$
Given $\lambda_1 = 5000$ \AA, $\lambda_2 = 6400$ \AA, and $D_1 = 100$ cm:
$D_2 = \dfrac{0.8 \times 5000 \times 100}{6400}$
$D_2 = \dfrac{400000}{6400} = 62.5$ cm
The displacement of the screen is $\Delta D = D_1 - D_2 = 100 - 62.5 = 37.5$ cm.
Since $D_2 < D_1$, the screen should be moved towards the source by $37.5$ cm.
Answer: towards the source by $37.5$ cm
The final fringe width is given by $\beta_2 = \dfrac{\lambda_2 D_2}{d}$
It is given that the fringe width is reduced by $20\%$ of its initial value, so:
$\beta_2 = \beta_1 - 0.2\beta_1 = 0.8\beta_1$
Substituting the expressions for fringe widths:
$\dfrac{\lambda_2 D_2}{d} = 0.8 \dfrac{\lambda_1 D_1}{d}$
$D_2 = \dfrac{0.8 \lambda_1 D_1}{\lambda_2}$
Given $\lambda_1 = 5000$ \AA, $\lambda_2 = 6400$ \AA, and $D_1 = 100$ cm:
$D_2 = \dfrac{0.8 \times 5000 \times 100}{6400}$
$D_2 = \dfrac{400000}{6400} = 62.5$ cm
The displacement of the screen is $\Delta D = D_1 - D_2 = 100 - 62.5 = 37.5$ cm.
Since $D_2 < D_1$, the screen should be moved towards the source by $37.5$ cm.
Answer: towards the source by $37.5$ cm
Hence, the correct answer is option: 'A'.