2018-10-01_Physics_For_You

e radius of curvature of the lens is 22 cm.

15. Let



E be in the x-direction and



B in the y-direction.
(a) Consider a rectangular loop in the x-z plane with

one side of length l parallel to



E. Suppose at any instant,
the rectangle is partially on the le of the wavefront and
partially on the right of the wave front.
Rate of change of magnetic ux,
d
dt

Blc

φ

=

Line integral of



EE=⋅∫ dl=El

From Faraday's law,

Edl

d

dt

⋅=B

∫

φ

? El = Blc or E = cB
(b) Consider a similar loop in the y-z plane

Rate of change of electric ux,

d

dt

E Elc

φ

=

Line integral of



BB=⋅∫ dl=Bl

From Ampere's law,



Bdl

d

dt

∫ ⋅=με E= Elc

φ
00 με 00
? Bl = P 0 H 0 Elc
or B = P 0 H 0 Ec = P 0 H 0 cB.c [ E = cB]

or cc^2

(^0000)

==^11

με με

or

16. Given gure shows the refraction of a plane
    wavefront at a rarer medium i.e., v 2 > v 1

i

i r

r

A

B

C

D

vt 1

vt 2

vv 12 <

Rarer–v 2

Denser –v 1

Incident

wavefront

Refracted
wavefront
e incident and refracted wavfronts are shown in
gure.
Let the angles of incidence and refraction be i and r
respectively.
From right 'ABC, we have,

sin ‘BAC = sin i =

BC

AC

From right 'ADC, we have,

sin ‘DCA = sin r = AD
AC

?

sin

sin

i

r

BC

AD

vt

vt

==^1

2

or

sin

sin

i

r

v

v

==^1

2

1

μ 2 (a constant)

is veries Snell’s law of refraction. e constant^1 P 2 is

called the refractive index of the second medium with

respect to rst medium.

17. (a) Angular width, q = λ
    d

or d = λ

θ

Here, O = 600 nm = 6 × 10–7 m

q = 0.1° =^01

180

. ×π rad = π
1800

rad, d =?

? d =^6101800

××−^7

π

= 3.44 × 10–4 m

(b) Frequency of a light depends on its source only. So,

the frequencies of reected and refracted light will be

same as that of incident light.

Reected light is in the same medium (air) so its

wavelength remains same as 500 Å.

Wavelength of refracted light, Or = λ

μw

Pw = refractive index of water.

So, wavelength of refracted wave will be decreased.

18. Maxwell’s generalization of Ampere’s circuital L aw,
     
    Bdliii

d

d dt

. =+()=+ E





∫ 

μμ 00 ε 0 φ

In the process of charging the capacitor there is change

in electric ux between the capacitor plates.

d

dt

d

dt

E EA

φ

= ()

E o Electric eld between the plates = q

Aε 0

A o Area of the plate

So,^

d

dt

d

dt

q

A

A dq

dt

φE id

εεε

=×









==

000

1

? id = i = H 0 d

dt

φE

19. Here E 0 = 120 NC–1,
    X = 50.0 MHz = 50 × 10^6 Hz

(a) B

E

(^0) c
0
1
81
(^1207)
310

== 410

×

=×

−

−

NC −

ms

T

Z = 2pX = 2 × 3.14 × 50 × 10^6

= 3.14 × 10^8 rad s–1.

k

c

==

×

×

= −

ω 31410

310

105

8

8

. .m 1