In a biprism experiment a steady interference pattern is observed on the screen using a light of wavelength 5000 A ˚ . Without disturbing the set up of the experiment, the source of light is replaced

Question

In a biprism experiment a steady interference pattern is observed on the screen using a light of wavelength 5000 A ˚ . Without disturbing the set up of the experiment, the source of light is replaced by a source of wavelength 6400 A ˚ . The fringe width will

TestHub · Accepted Answer

The fringe width in a biprism experiment is given by $X = \frac{λ D}{d}$ .

Since the experimental setup is not disturbed, the distance between the source and screen $D$ and the distance between the coherent sources $d$ remain constant. Therefore, $X \propto λ$ .

Taking the ratio of the new fringe width to the initial fringe width:

$\frac{X _{2}}{X _{1}} = \frac{λ _{2}}{λ _{1}}$

$\frac{X _{2}}{X _{1}} = \frac{6400}{5000} = 1.28$

$X_{2} = 1.28 X_{1}$

The percentage change in the fringe width is:

$\frac{X _{2} - X _{1}}{X _{1}} \times 100 = \frac{1.28 X _{1} - X _{1}}{X _{1}} \times 100 = 0.28 \times 100 = 28%$

Since the value is positive, the fringe width increases by $28%$ .

Answer: increase by 28 %

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