Question

In young’s double slit experiment, the fringe width of a width light of wavelength 6000Å is 3 mm. the fringe width, when the wavelength of light is changed to 4000Å is :

• A. 3 mm
• B. 1 mm
• C. 2 mm
• D. 4 mm

Answer 1

Answer: (C)

2 mm

Answer 2

Fringe width $\beta = \frac{\lambda D}{d}$

Where $\lambda$is the wavelength of light D is distance between slits and the screen d is distance between the two slits. As D and d remain the same $\beta \propto \lambda$

Or $\frac{\beta'}{\beta} = \frac{\lambda'}{\lambda}or\beta' = \frac{\lambda'B}{\lambda}$

Substituting the given values, we get

$\beta' = \frac{4000Å \times 3mm}{6000Å} = 2mm$