In a biprism experiment, the slit and the eyepiece. are 10 cm and 80 c

1.	In a biprism experiment, the slit and the eyepiece. are 10 cm and 80 cm away from the biprism. When a convex lens was interposed at 30 cm from the slit, the separation of the two magnified images of the slit was found to be 4.5 mm. If the wavelength of the source is 4500 Å, calculate the fringe width.
Answer» Data : d₁= 4.5 mm = 4.5 × 10^-3 m, λ = 4500 Å = 4.5 × 10^-7 m, distance between the slit and the eyepiec = distance between the slit and the biprism + distance between the biprism and the eyepiece = 10 cm + 80 cm = 90 cm = 0.9 m, u₁ = 30 cm = 0.3 m v₁ = D – u₁ = 0.9 m – 0.3 m = 0.6 m Linear magnification of a lens, \(\cfrac{image\,size}{object\,size}\) = \(\cfrac{image\,distance}{object\,distance}\) \(\therefore\) \(\cfrac{d_1}d\) = \(\cfrac{v_1}{u_1}\) \(\therefore\) d = \(\cfrac{d_1u_1}{v_1}\) = \(\cfrac{4.5\times10^{-3}\times0.3}{0.6}\) \(\therefore\) d = 2.25 x 10^-3 m \(\therefore\) The fringe width, W = \(\cfrac{\lambda D}d\) = \(\cfrac{4.5\times10^{-7}\times0.9}{2.5\times10^{-3}}\) = 2 x 0.9 x 10^-4 = 1.8 x 10^-4 m = 0.18 mm

In a biprism experiment, the slit and the eyepiece. are 10 cm and 80 cm away from the biprism. When a convex lens was interposed at 30 cm from the slit, the separation of the two magnified images of the slit was found to be 4.5 mm. If the wavelength of the source is 4500 Å, calculate the fringe width.

Answer»

Data : d₁= 4.5 mm = 4.5 × 10^-3 m,

λ = 4500 Å = 4.5 × 10^-7 m,

distance between the slit and the eyepiec = distance between the slit and the biprism + distance between the biprism and the eyepiece

= 10 cm + 80 cm = 90 cm = 0.9 m,

u₁ = 30 cm = 0.3 m

v₁ = D – u₁

= 0.9 m – 0.3 m

= 0.6 m

Linear magnification of a lens,

\(\cfrac{image\,size}{object\,size}\) = \(\cfrac{image\,distance}{object\,distance}\)

\(\therefore\) \(\cfrac{d_1}d\) = \(\cfrac{v_1}{u_1}\)

\(\therefore\) d = \(\cfrac{d_1u_1}{v_1}\) = \(\cfrac{4.5\times10^{-3}\times0.3}{0.6}\)

\(\therefore\) d = 2.25 x 10^-3 m

\(\therefore\) The fringe width,

W = \(\cfrac{\lambda D}d\) = \(\cfrac{4.5\times10^{-7}\times0.9}{2.5\times10^{-3}}\) = 2 x 0.9 x 10^-4