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
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 = ?
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,
(1)
(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,
(3)
By inserting the values in equation (3) we get;
(4)
Now by putting the values in equation (2) we get;