(b) Advantage of reecting telescope, over a refracting

type:

(i) In refracting telescope the nal image is formed

aer two times of partial refraction through the lenses,

major losses in the intensity take place due to partial

reection and refractions. In reecting telescope, all the

light intensity incident forms the nal image as no loss

of intensity can be ensured in reection.

(ii) In refracting telescope glass of lens oers dierent

refractive indices to dierent colours and hence

chromatic abberation due to which coloured image

is formed take place. Reecting telescope is free from

chromatic abberation.

24. (a) Sun glasses lled with polaroid sheets protect
    our eyes from glare. Polaroids reduce head light glare of
    motor car being driven at night. Polaroids are used in
    three dimensional pictures, i.e., in holography.
    (b) Let two polaroids P 1 and P 2 are placed in crossed
    positions. Let P 3 be the polaroid sheet placed between
    P 1 and P 2 making an angle q with pass axis of P 1
    If I 1 = intensity of polarised light aer passing through
    P 1 then intensity of light aer passing through P 3 will be
    I = I 1 cos^2 q ...(i)
    Now angle between P 2 and P 3

=−









π

θ

2

[... P 1 and P 2 are in crossed position]

? Outcoming intensity aer P 2 is,

I 2 = I cos^2 (90° – q)

or I 2 = I 1 cos^2 q sin^2 q = 






I 1

1 2

2

sin 2 θ

p/4 p/2 3/p 4 p q

I 2

25. (a) Refer to answer 60(a), Page no.225 (MTG CBSE
    Champion Physics Class 12)
    (b) R = 20 cm, n 2 = 1.5, n 1 = 1, u = –100 cm
    n
    v

n

u

nn

R

(^21) −= 21 − or^151
100

151

20

..

v

+=−

or

15 05

20

1

100

1

40

1

100

3

200

..

v

=−=−=

or v =

200

3

× 1.5 = 100 cm

Ÿ So, a real image is formed on the other side,

100 cm away from the surface.

OR

Refer to answer 73, Page no. 264 (MTG CBSE Champion

Physics Class 12)

26. (a) Refer to answer 85(a), Page no. 228 (MTG CBSE
    Champion Physics Class 12)
    (b) Critical angle for

(i) Red light is sincr==

1

139

0 7194

.

.

or cr = 46°

(ii) Green light is sincg==

1

144

0 6944

.

.

or cg = 44°

(iii) Blue light is sincb =

1

14 7.

= 0.6802

or cb = 43°

As angle of incidence i = 45° of red light ray on face AC

is less that its critical angle of 46°, so red light ray will

emerge out of face AC.

A

B C

B

G

R

B G

R

OR

Refer to answer 38, Page no. 257 (MTG CBSE Champion

Physics Class 12)

27. (a) Refer to answer 130, Page no. 236 (MTG CBSE
    Champion Physics Class 12)
    OR
    (a) Refer to answer 104 (a), Page no. 270 (MTG CBSE
    Champion Physics Class 12)

(b) Intensity observed by observer O 1 = I^0

2

Intensity observed by observer

O 2 =

I 0 260 °

2

cos =×II (^00) =
2
1
48
Intensity observed by observer O 3
=°−°
I 0 2
8
cos( 90 60 )=°
I 0 2
8
sin 30
=×




II 0 2 0
8
1
232