Physics, asked by panchalrenuka1, 4 months ago

83. In a biprism experiment, the slit and
eye-piece are 15 cm and 40 cm away from the
biprism. When a convex lens interposed at
20 cm from the slit, the separation of two
magnified images of the slit is found to be
4.5 mm. If the wavelength of the source is
4200 Å, calculate the fringe width.
ution:​

Answers

Answered by vk5528552
2

Answer:

When a convex lens interposed at 20 cm from the slit, the separation of two magnified images of the slit is found to be 4.5 mm. ... Solution: Given: di = 4.5 mm = 4.5 x 10 m, 1 = 4200 A = 4.2 x 10'm, D = 15 + 40 = 55 cm = 0.55 m, u = 20 cm = 0.2 m, Vi =D-u=0.55-0.2 = 0.35 m To find: X = ?

Answered by archanajhaasl
4

Answer:

Explanation:

From the question we have,

The separation of two magnified images of the slit(d₁)=4.5mm=4.5×10⁻³m

The wavelength of the source(λ)=4200A°=4.2×10⁻⁷m

The distance between slit and eye-piece=(15+40)cm=55cm=0.55m

The distance of the slit from the convex lens(u₁)=20 cm=0.2m

v₁=D-u₁=0.55-0.2=0.35m

The formula that we are going to use is,

\mathrm{\frac{d_1}{d} =\frac{v_1}{u_1} }              (1)

\mathrm{W=\frac{\lambda D}{d} }            (2)

Where,

d=size of object

d₁=size of image

v₁=image distance==D-u₁=0.55-0.2=0.35m

u₁=object distance

W=fringe width

λ=wavelength of light used

D=distance between slit and the eye-piece

We can modify equation (1) as,

\mathrm{d=\frac{d_1u_1}{v_1} }          (3)

By inserting the values in equation (3) we get;

\mathrm{d=\frac{4.5\times 10^{-3}\times 0.2}{0.35} }

\mathrm{d=2.57\times 10^{-3}m}      (4)

Now by putting the values in equation (2) we get;

\mathrm{W=\frac{4.2\times 10^{-7}\times  0.55}{2.57\times 10^{-3}} }

\mathrm{W=8.988\times 10^{-5}m}

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