λ

υ

==

×

×

=

c 310

50 10

600

8

6 .m

(b) If the wave is propagating along x-axis, then eld



E

will be along y-axis and eld



B along z-axis.

?

 

EE=− 0 sin(kx ωtj)

or

 

Ex=− 120 sin( 10 .. 53 14 10×^81 tj)NC−

where x is in metre and t in second.

 

BB=− 0 sin(kx ωtk)

= 41 ×− 01 −^78 sin(.. 05 xt 3141 × 0 )ktesla

20. As per the gure,
    e image formed by lens L 1 is at P. erefore, using

lens formula^111

fvu

=−

As per the parameters given in the question

u = – 15 cm, fL 1 = 20 cm

So, the image distance will be

11

15

1

v 20

−

−

=
()
v = – 60 cm
Now, this image is acting as an object for the lens L 2.
We can again use the lens formula and other parameters
given in the question and question gure to nd the
focal length of lens L 2.
111
vuLL 22 fL 2

−=

Here, uL 2 = v + (–20) = – 60 – 20 = – 80 cm

vL 2 = 80 cm

1

80

1

80

1

2

−

−

=

()fL

fL 2 = 40 cm

So, the focal length of the lens L 2 = 40 cm.

21. Here, height of object h = 3 cm
    u = –60 cm, f = +30 cm
    Using the mirror formula, we have
    111
    vuf

+=

11

60

1

30

11

30

1

vv 60

+

−

=⇒=+

121

60

13

vv 60

= + ⇒=? v = 20 cm

e image is virtual and erect.
e image is at a distance of 20 cm from the mirror on
the opposite side of the object.

′

=− ⇒

′

=−

−







 ⇒

′

=

h

h

v

u

hh

3

20

60 3

1

3

Ÿ hc = 1 cm

? Image is diminished and its size is 1 cm.

OR

(a) (i)

X

Y

Z

Er

r

B

(ii) Speed of e.m. wave can be given as the ratio of

magnitude of electric eld (E 0 ) to the magnitude of

magnetic eld (B 0 ), i.e., c

E

=B

0

0

(b) Electromagnetic waves or photons transport energy

and momentum. When an electromagnetic wave

interacts with a small particle, it can exchange energy

and momentum with the particle. e force exerted on

the particle is equal to the momentum transferred per

unit time. Optical tweezers use this force to provide a

non-invasive technique for manipulating microscopic-

sized particles with light.

22. Given that: Wavelength of the light beam,
    O 1 = 590 nm = 5.9 ×10–7 m
    Wavelength of another light beam,
    O 2 = 596 nm = 5.96 × 10–7 m
    Distance of the slits from the screen = D = 1.5 m
    Slits width = a = 2 × 10–4 m
    For the rst secondary maxima,

sin q ==

3

2

λ 11

a

x

D

x

D

(^1) a
(^31)
2

λ and
x
D
(^2) a
(^32)
2

λ
? Separation between the positions of rst secondary
maxima of two sodium lines,
xx D
(^21) a 21
3
2
−=()λλ−
= ×
××−
315
22104
.
(5.96 × 10–7 – 5.9 × 10–7) = 6.75 × 10–5 m

23. (a) e formation of image by a reecting telescope,
    Objective mirror

Eye piece

Secondary

mirror