Question

Yellow light of wavelength 6000a produces fringes of width 0.8 mm in young double slit experiment what will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500a and the separation between the slit is double

Answer 1

solution is attached

Answer 2

Answer:

The fringe width due to second light is 0.5 mm

Explanation:

The wavelength of the light is $\lambda_{1}=6000 A$

The fringe width is $\beta_{1}=0.8\ mm$

The distance between the slit is d

Now the light is replaced

The wave length of the second light is $\lambda_{2}=7500\ A$

The distance between the slit is doubled d'=2d

We know the relation of fringe width

$\beta=\dfrac{\lambda D}{d}$

The formula for fringe width for the first light is

$\beta_{1}=\dfrac{\lambda_{1} D}{d}$           (1)

The formula for for fringe width for the second light is,

$\beta_{2}=\dfrac{\lambda_{2} D}{2d}$         (2)

Dividing equation 2 to 1 we get,

$\dfrac{\beta_{2}}{\beta_{1}}=\dfrac{\lambda_{2}}{\lambda_{1}}.\dfrac{d}{2d}\\\beta_{2}=\beta_{1}\dfrac{\lambda_{2}}{\lambda_{1}}.\dfrac{1}{2}\\\beta_{2}=0.8\times \dfrac{7500}{6000}.\dfrac{1}{2}\\\beta_{2}=0.5\ mm$