1. Xtra Edge February 2012 - Career Point

4. In a Young's experiment, the upper slit is covered by a
thin glass plate of refractive index µ1 = 1.4 while the
lower slit is covered by another glass plate having the
same thickness as the first one but having refractive
index µ2 = 1.7. Interference pattern is observed using
light of wavelength λ = 5400 Å. It is found that the
point P on the screen where the central maxima fell
before the glass plates ware inserted, now has 3/4 the
original intensity. It is further observed that what used
to be the fifth maxima earlier, lies below the point P
while the sixth minima lies above the point P.
Neglecting absorption of light by glass plates, calculate
thickness of the glass plates.
Sol. If glass plates are not inserted in slits, central
maximum occurs at perpendicular bisector of slits.
Hence, point P lies on perpendicular bisector of S1S2
as shown in figure.
S1
S2
Screen
When a plate is inserted in upper slit, the
interference pattern shifts upwards and due to plate
inserted in lower slit, it shifts downwards.
Since, distance of nth maxima (nth bright fringe)
from central bright fringe is nω, where ω is fringe
width and fifth maxima now lies below the point P,
therefore, shift of fringe pattern is downwards and it
is greater that 5 ω.
Since, sixth minima (sixth dark fringe) lies above
point P and distance of m th minima from central
⎛ 1 ⎞
bright fringe is ⎜m
– ⎟ ω, therefore, shift of fringe
⎝ 2 ⎠
pattern is less than 5.5 ω. Hence, the shift lies
between 5 ω and 5.5 ω.
Let intensity of light due to each slit be I0. Then
l1 = l2 = l0.
Since, before insertion of glass plates, a bright fringe
(central bright fringe) was formed at P, therefore,
original intensity at P was equal to Imax.
But Imax = ( 1 I + I 2 ) 2 = 4I0
But now intensity at P is 3/4 of the original intensity,
therefore now intensity at P is
I = 3/4 Imax = 3I0.
But I = I1 + I2 + 1 2 I I 2 cos φ where φ is phase
difference between two rays reaching P.
Substituting I1 = I2 = I0 and I = 3I0 in above
equation,
1
cos φ = or φ = (2nπ + π/3) where n is an integer
2
P
s
But phase difference, φ = 2π where s is shift of
ω
fringe pattern.
⎛ 1 ⎞
Hence, s = ⎜n
+ ⎟ ω.
⎝ 6 ⎠
But s lies between 5ω and 5.5 ω, therefore,
⎛ 1 ⎞ 31 λ D
s = ⎜5
+ ⎟ ω = ω where ω =
⎝ 6 ⎠ 6
d
31λ D
∴ s = where D is distance of screen from
6d
slits and d is distance between slits.
Let thickness of each glass plate be t.
∴ Upward shift due to upper glass plate,
( µ 1 – 1)
tD
s1 =
d
and downward shift due to lower glass plate,
( µ 2 – 1)
tD
s2 =
d
∴ Resultant downward shift s = s2 – s1.
Substituting values of s, s1 and s2,
31λ
t =
6(
µ –
= 9.3 µm
)
Ans.
2 µ 1
5. A point isotropic light source of power P = 12 watt is
located on the axis of a circular mirror plate of radius
R = 3 cm. If distance of source from the plate is
a = 39 cm and reflection coefficient of mirror plate is
α = 0.70, calculate force exerted by light rays on the plate.
Sol. When light rays are incident on the mirror plate, a
part of then is reflected and a part is absorbed by the
plate. Therefore, momentum of light rays changes.
Due to change in the momentum, a force is
experienced by the plate. Magnitude of the force is
equal to rate of change of momentum. To calculate
rate of change of momentum, first we have to
consider a circular ring coplanar and co-axial with
the plate. Let the radius of that ring be x and radial
thickness dx as shown figure
Source
XtraEdge for IIT-JEE 19 FEBRUARY 2012
a
Distance of every point of this ring from the source
is
2 2
r = a + x .
Hence, intensity of light incident on the ring is
P
I =
2
4πr
θ