2018-10-01_Physics_For_You

Ÿ sin()

sin( )

r =

°

=

60

3

1

2

Ÿ r=sin−^1  

1
2
? r = 30°
(b) Focal length of the lens decrease when red light is
replaced by blue light.
OR
(a) e essential condition, which must satised
sources to be coherent are :
(i) the two light waves should be of same wavelength.
(ii) the two light waves should either be in phase or
should have a constant phase dierence.
(b) Because coherent sources emit light waves of same
frequency or wavelength and of a stable phase dierence.

12. (a) Essential conditions for total internal reection :
    (i) Light should travel from a denser medium to a rarer
    medium.
    (ii) Angle of incidence in denser medium should be
    greater than the critical angle for the pair of media in
    contact.
    (b) (i) To deviate a ray of light through 90° :

90°

A

B Q C

P R

45°

45° 45°

A totally reecting prism is used to deviate the path

of the ray of light through 90°, when it is inconvenient

to view the direct light. In Michelson’s method to nd

velocity of light, the direct light from the octagonal

mirror is avoided from direct viewing by making use of

totally reecting prism.

(ii) To deviate a ray of light through 180° : When the ray

of light comes to meet the hypotenuse face BC at right

angles to it, it is refracted out of prism as such along the

path RS. e path of the ray of light has been turned

through 180° due to two total internal reections.

90°

45°

A

B C

Q R

P S

45°

45° 45°

13. (a) The refractive index of the material of prism,

P =

sin

sin

A

A

 + m







δ

2

2

Given : A = 60°, Gm = 30°

P =

sin

sin

45

30

1

2

°

°

=. 2 Ÿ P = 2

' P =

c

v

Ÿ v =

c

μ

=

31 × 0

1 414

8

.

= 2.12 × 10^8 m s–1

(b) sin iC =^11

μ 2

=^

r 1 r

δm

A

B C

i

iC = r = 45°

A = r 1 + r

Ÿ r 1 = 15°

sin

sin

i

r 1

= 2

sin i = 2 sin 15° =

() 31

22

2

−

×

sin i =

31

2

−

i = sin–1

31

2

 −











14. (a) e focal length of original equiconvex lens is f.
    Let the focal length of each part aer cutting be F.
    Here,

11112

fFFfF

=+⇒=

⇒=f F⇒=Ff

2

2

Power of each part will be given by

P

F

P

f

=⇒^11 =

2

(b) From lens maker formula, we have

P

RR

=− −









()μ 1

11

12

where

P = Power of lens = +5 D

P = Refractive index of the lens = 1.55

R 1 = Radius of curvature of rst face (+ve)

R 2 = Radius of curvature of second face (–ve)

Given : R 1 = R 2 = R

⇒= − −

−









⇒= − 







51 (. 55 1 )(^1151 .) 55 1 2

RR R

⇒= ⇒=

×

5055 2 ⇒=

0552

5

. 022

.

.

R

RRm