1000 Solved Problems in Modern Physics

6.3 Solutions 341

and at the counter C

Ic=IAexp (− 2 d/γβcτ)(2)

∴Ic=

IB^2

IA

=

(470)^2

1000

= 221

(b) Take logarithm on both sides of (1) and simplify.

τ=

d

γβcln(IA/IB)

(3)

γ= 1 +T/mc^2 = 1 + 140 / 140 = 2. 0

β=(γ^2 −1)^1 /^2 /γ= 0. 866

d=10 m;c= 3 × 108 m/s

ln(IA/IB)=ln(1000/470)= 0. 755

Substituting the above values in (3),

τ= 2. 55 × 10 −^8 s

The accepted value is 2. 6 × 10 −^8 s

6.36 The stationary object will appear to move with velocity−βctoward the
observer. The object moving with velocityαctoward the stationary object
would appear to have velocity
(αc−βc)/(1−αβ), as seen by the observer. If these two velocities are to be
equal then (αc−βc)/(1−αβ)=βc
Cross multiplying and simplifying we get the quadratic equation whose
solution isβ=[1−(1−α^2 )^1 /^2 ]/α

6.37
(i) t=

L

βc

(1)

N 2 =N 1 exp[−t/γτ]=N 1 exp

[

−

L

γβcτ

]

(2)

Therefore (2) becomes

exp

[

L

γβcτ

]

=N 1 /N 2

Take logarithm on both sides

L

γβcτ=ln

(

N 1

N 2

)

Butγβ=

√

γ^2 − 1

Thereforeτ= L

ln

(N

N^1

2

)√

γ^2 − 1 c

(ii) γ= 1 /(1−β^2 )^1 /^2 = 1 /(1− 8 /9)^1 /^2 = 3

τ=

200

ln

(

10 , 000

8 , 983

)√

32 − 1 × 3 × 108

= 2. 2 × 10 −^6 s.