Question

A beam of light consisting of two wavelength 6500A∘ and 5200A∘ is used to obtain interference fringes in a young's double slit exp. Then what is the distance from central maxima where the bright fringes due to both wavelength coincides? The distance between slits is 2 mm and distance between plane of slits and screen is 120 cm

Answer 1

Answer:

ANSWER

LCMLCM of 6565 and 52=26052=260

i.e., \lambda =26000\mathring { A }λ=26000

A

˚

\beta =\cfrac { \lambda D }{ d } =\cfrac { 26000 }{ { 10 }^{ 10 } } \times \cfrac { 120 }{ 100 } \times \cfrac { 1000 }{ 2 }β=

d

λD

=

10

10

26000

×

100

120

×

2

1000

=\cfrac { 156 }{ { 10 }^{ 5 } }=

10

5

156

m

=1.56=1.56mm

Answer 2

The distance from central maxima where the bright fringes due to both wavelength coincides is :

• Given : The distance between slits, d = 2 mm

and distance between plane of slits and screen, D = 120 cm

λ1 = 6500Å, λ2 = 5200Å

• LCM of 6500 and 520p = 26000

• λ = 26000Å is the wavelength for which both bright fringes formed by given wavelength coincides.

• We know that,

β = λD/d

= 26000×120×1000 / 10^10×100×2

= 156 / 10^5

= 1.56 mm