# Chapter 21 Refraction Of Light¶

## Example 21.1 Page no 650¶

In [6]:
#Given
U=1.5
c=3.0*10**8                          #m s**-1

#Calculation
v=c/U

#Result
print" Speed of light in glass is",v*10**-8,"10**8","m s**-1"

 Speed of light in glass is 2.0 10**8 m s**-1


## Example 21.2 Page no 650¶

In [35]:
#Given
u=1.33
Lu=589*10**-9                             #m
c=3*10**8

#Calculation
v=c/Lu
v1=c/u
Lw=v1/v

#Result
print"(a) Wavelenght of light after reflection is",Lu*10**9,"10**-9"
print" Velocity of light after reflection is",c*10**-8,"10**8","m s**-1"
print" Frqequency of light after reflection is",round(v*10**-14,4),"10**14","Hz"
print"(b) Frequency of light after refraction is",round(v*10**-14,4),"10**14","Hz"
print" Velocity of light after refraction is",round(v1*10**-8,3),"10**8","m s**-1"
print" Wavelength of light after refraction is",round(Lw*10**9,2),"nm"

(a) Wavelenght of light after reflection is 589.0 10**-9
Velocity of light after reflection is 3.0 10**8 m s**-1
Frqequency of light after reflection is 5.0934 10**14 Hz
(b) Frequency of light after refraction is 5.0934 10**14 Hz
Velocity of light after refraction is 2.256 10**8 m s**-1
Wavelength of light after refraction is 442.86 nm


## Example 21.3 Page no 651¶

In [48]:
#Given
i=60
u=1.5
t=0.1                          #m

#Calculation
import math
r=(math.sin(60*3.14/180.0))/u
a=math.asin(r)*180/3.14
d=t/math.cos(r*3.14/180.0)*(math.sin(24*3.14/180.0))

#Result
print"The lateral shift produced is",round(d,4),"m"

The lateral shift produced is 0.0407 m


## Example 21.4 Page no 651¶

In [59]:
#Given
A=4/3.0
A1=3/2.0
I=30

#Calculation
import math
S=A1/A
R=math.sin(30*3.16/180.0)/S
A=math.asin(R)*180/3.14

#Result
print"The angle of refraction is",round(A,0)

The angle of refraction is 27.0


## Example 21.5 Page no 651¶

In [67]:
#Given
t=15
u=1.5

#Calculation
d=t*(1-(1/u))

#Result
print" Real thickness of glass slab is",d,"cm"
print" The answer does not depend upon the location of the slab"

 Real thickness of glass slab is 5.0 cm
The answer does not depend upon the location of the slab


## Example 21.6 Page no 651¶

In [73]:
#Given
c=3*10**8                         #m s**-1
v=2.0*10**8                         #m s**-1
t=6.0                              #cm

#Calculation
u=c/v
d=t*(1-(1/u))

#Result
print"An ink dot appear to be rasied",d,"cm"

An ink dot appear to be rasied 2.0 cm


## Example 21.7 Page no 651¶

In [84]:
#Given
d=12.5                          #cm
a=9.4                           #cm
u=1.63

#Calculation
S=d/a
S1=d/u
S3=a-S1

#Result
print"The refractive index of water is",round(S,2)
print"The refractive index of liquid is",round(S1,2),"cm"
print"Distance is",round(S3,2),"cm"

The refractive index of water is 1.33
The refractive index of liquid is 7.67 cm
Distance is 1.73 cm


## Example 21.8 Page no 651¶

In [91]:
#Given
A=4/3.0
d=0.015                            #m

#Calculatiom
t=d/(1-(1/A))

#Result
print"The height upto which water must be poured into the beaker is",t,"m"

The height upto which water must be poured into the beaker is 0.06 m


## Example 21.9 Page no 652¶

In [102]:
#Given
A=3/2.0
A1=4/3.0
t1=6
t2=4

#Calculation
d1=t1*(1-(1/A))
d2=t2*(1-(1/A1))
d3=d1+d2

#Result
print"The apparent position of an object is",d3,"cm"

The apparent position of an object is 3.0 cm


## Example 21.10 Page no 652¶

In [124]:
#Given
S=1.5
W=1.33

#Calculation
import math
A=S/W
C=1/A
Q=math.asin(C)*180/3.14

#Result
print"The critical angle for a glass water interface is",round(Q,0),"degree"

The critical angle for a glass water interface is 62.0 degree


## Example 21.11 Page no 652¶

In [123]:
#Given
I=40                             #degree
A=15                             #degree

#Calculation
import math
r=I-A
u=(math.sin(40*3.14/180.0))/(math.sin(25*3.14/180.0))
c=1/u
A=math.asin(c)*180/3.14

#Result
print"The critical angle is",round(A,1),"degree"

The critical angle is 41.1 degree


## Example 21.12 Page no 652¶

In [138]:
#Given
u=1.5
h=20

#Calculation
import math
C=1/u
A=math.asin(C)*180/3.14
r=h*math.tan(A*3.14/180.0)
D=math.pi*r**2

#Result
print"The required surface area is",round(D,1),"cm**2"

The required surface area is 1005.3 cm**2


## Example 21.13 Page no 653¶

In [143]:
#Given
A=60                                  #degree
B=1.45

#Calculation
import math
I=1/(math.sin(A*3.14/180.0))
Z=B/I

#Result
print"The refractive index of the liquid is",round(Z,3)

The refractive index of the liquid is 1.255


## Example 21.14 Page no 653¶

In [161]:
#Given
A=1.68
A1=1.44
A2=90                               #degree

#Calculation
import math
K=A/A1
C=1/K
Q=math.asin(C)*180/3.14
r=A2-Q
I=A*math.sin(r*3.14/180.0)
Q1=math.asin(I)*180/3.14

#Result
print"Refraction index of glass fibre is",round(Q1,0),"degree"
print"All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection"

Refraction index of glass fibre is 60.0 degree
All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection


## Example 21.15 Page no 653¶

In [178]:
#Given
u1=1
u2=1.5
u=-10.0                           #cm
v=-40.0                          #cm

#Calculation
R=(u2-u1)/((-u1/u)+(u2/v))

#Result

Radius of curvature is 8.0 cm


## Example 21.16 Page no 653¶

In [191]:
#Given
u=-60.0                             #cm
R=25.0                              #cm
u1=1
u2=1.5
R1=0.25

#Calculation
v=u2/(((u2-u1)/R)+(u1/u))
P=(u2-u1)/R1

#Result
print"Position of the image is", v,"cm"
print"Power of refracting surface is",P,"dioptre"

Position of the image is 450.0 cm
Power of refracting surface is 2.0 dioptre


## Example 21.17 Page no 654¶

In [194]:
#Given
U1=1.5
U2=1

#Calculation
x=(U1+U2)/(U1-U2)

#Result
print"Distance of the object is",x,"R"

Distance of the object is 5.0 R


## Example 21.18 Page no 654¶

In [196]:
#Given
u1=1
u2=1.5
u=-10.0                           #cm
R=-5.0

#Calculation
v=u1/((u1-u2)/R+(u2/u))

#Result
print"Position of the image is",v,"cm"

Position of the image is -20.0 cm


## Example 21.19 Page no 654¶

In [202]:
#Given
u=1.5
R=-5.0                              #cm
OC=2

#Calculation
u1=R+OC
v=1/((1-u)/R+(u/u1))

#Result
print"Apparent position of the bubble is", v,"cm"

Apparent position of the bubble is -2.5 cm


## Example 21.20 Page no 654¶

In [206]:
#Given
R=-0.02                            #m
u=1.54
v=-0.01                             #m

#calculation
u1=-u/((1-u)/R-(1/v))

#Result
print"Real depth of bubble is", round(u1,4),"m"

Real depth of bubble is -0.0121 m


## Example 21.21 Page no 655¶

In [213]:
#Given
R=7.5                       #cm
u=4/3.0

#Calculation
v=1/((1-u)/R)

#Result
print"It get focussed at",v,"cm"

It get focussed at -22.5 cm


## Example 21.22 Page no 655¶

In [215]:
#given
u=1.55
f=20                               #cm

#Calculation
R=(u-1)*2*f

#Result

Radius of curvature is, 22.0 cm


## Example 21.23 Page no 655¶

In [218]:
#Given
u=1.5
f=0.3

#Calculation
R=(u-1)*f

#Result

Radius of curvature is 0.15 m


## Example 21.24 Page no 655¶

In [219]:
#Given
a=2.0

#Calculation
u=(a+1)/a

#Result
print"Refractive index is",u

Refractive index is 1.5


## Example 21.25 Page no 655¶

In [225]:
#Given
ug=1.5
uc=1.63
R1=20.0                        #cm
R2=-20.0

#Calculation
fair=1/((ug-1)*(1/R1-1/R2))
ug1=ug/uc
fc=1/((ug1-1)*(1/R1-1/R2))

#Result
print"Focal length of the lens is",round(fc,1),"cm"

Focal length of the lens is -125.4 cm


## Example 21.26 Page no 656¶

In [234]:
#Given
fair=0.2                    #m
ug=1.50
uw=1.33

#Calculation
R=1/(fair*(ug-1))
ug1=ug/uw
fw=1/((ug1-1)*R)
f=fw-fair

#Result
print"Change in focal length is", round(f,2),"m"

Change in focal length is 0.58 m


## Example 21.27 Page no 656¶

In [237]:
#Given
u1=1.2
u2=1.3
a=13.0

#Calculation
u=u1/u2
f=1/((-1/a)*(1/(u1-1)))

#Result
print"Focal length is", f,"f"

Focal length is -2.6 f


## Example 21.28 Page no 656¶

In [244]:
#Given
f=20.0                    #cm\
u=12.0
f1=-20.0

#Calculation
v=1/(1/f+1/u)
v1=1/(1/f1+1/u)

#Result
print"(a) Beam will converge at a point of distant",v,"cm"
print"(b) Beam will converge at a point of distant",v1,"cm"

(a) Beam will converge at a point of distant 7.5 cm
(b) Beam will converge at a point of distant 30.0 cm


## Example 21.29 Page no 656¶

In [249]:
#Given
f=-0.2                           #m
v=0.3                            #m

#Calculation
u=1/(1/v-1/f)

#Result
print"The position of the point is",u,"m"

The position of the point is 0.12 m


## Example 21.30 Page no 656¶

In [273]:
#Given
f=10.0                                 #cm
c=20                                 #cm
o=5.0

#Calculation
u=-(c-o)
u1=-(c+o)
v1=1/((1/f)+(1/u))
v2=1/((1/f)+(1/u1))
v=v1-v2

#Result
print"Length of image of needle is", round(v,2),"cm"

Length of image of needle is 13.33 cm


## Example 21.31 Page no 657¶

In [289]:
#Given
u=-30.0
v=20.0
R1=10.0
R2=-15.0
ug1=1.33

#Calculation
f=1/(-1/u+1/v)
ug=(1/f+1/6.0)*6.0
fw=1/(((ug/ug1)-1)*(1/R1-1/R2))

#Result
print"Focal length of the lens is", f,"cm"
print"Focal length of the lens is",round(fw,2),"cm"

Focal length of the lens is 12.0 cm
Focal length of the lens is 46.94 cm


## Example 21.32 Page no 657¶

In [292]:
#Given
f=0.12
m=-3.0

#Calculation
u=(f/m)-f

#Result
print"Distance between object and lens is", u,"m"

Distance between object and lens is -0.16 m


## Example 21.33 Page no 657¶

In [296]:
#Given
m=-19.0
v=10

#Calculation
f=v/(-m+1)

#Result
print"Focal length of the lens is",f,"m"

Focal length of the lens is 0.5 m


## Example 21.34 Page no 657¶

In [301]:
#Given
m=4
f=20

#Calculation
u=m+1-f
v=f-(m*f)

#Result
print"Object is at", u,"cm"
print"Image is at",v,"cm"

Object is at -15 cm
Image is at -60 cm


## Example 21.35 Page no 657¶

In [309]:
#Given
O=5                      #cm
u=-45.0                       #cm
v=90.0

#Calculation
f=-1/(1/u-1/v)
I=v*O/u

#Result
print"Focal length is", f,"cm"
print"size of the image is",I,"cm"

Focal length is 30.0 cm
size of the image is -10.0 cm


## Example 21.36 Page no 658¶

In [3]:
#Given
O=3                                    #cm
u=-14.0
f=-21.0

#Calculation
v=1/(1/f+1/u)
I=O*v/u

#Result
print"Image produced by the lens is",I,"cm"

Image produced by the lens is 1.8 cm


## Example 21.37 Page no 658¶

In [8]:
#Given
R1=0.20
R2=-0.2
u=1.5
u1=1.25

#Calculation
P1=(u-1)*(1/R1-1/R2)
u2=u/u1
P2=(u2-1)*(-1/R2-1/R2)
P=P1/P2

#Result
print"Ratio of power of lens is", P

Ratio of power of lens is 2.5


## Example 21.38 Page no 658¶

In [16]:
#Given
f1=15.0                   #cm
f2=30.0

#Calculation
f=1/(1/f1+1/f2)
P=1/f

#Result
print"Power is",f*10**-2,"m"
print"Focal length is",P*10**2,"D"

Power is 0.1 m
Focal length is 10.0 D


## Example 21.39 Page no 658¶

In [18]:
#Given
P1=-1.5                        #D
P2=2.75

#Calculation
P=P1+P2
f=1/P

#Result
print"Focal length is",f,"m"

Focal length is 0.8 m


## Example 21.40 Page no 659¶

In [24]:
#Given
a=0.11
a1=0.07

#Calculation
x=a-a1
u=(a1-0.01)/x

#Result
print"Refractive index is",u

Refractive index is 1.5


## Example 21.41 Page no 659¶

In [29]:
#Given
a=70                                #Degree

#Calculation
import math
a=1/(math.sin(a*3.14/180.0))
a1=math.atan(a)*180/3.14
u=1/math.sin(a1*3.14/180.0)

#Result
print"Critical angle is", round(u,3)

Critical angle is 1.372


## Example 21.42 Page no 659¶

In [34]:
#Given
a=180                          #degree
b=90
c=45

#Calculation
import math
r=a-(b+c)
A=1.352*math.sin(r*3.14/180.0)
A1=math.asin(A)*180/3.14

#Result
print"(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]"
print"(b) Angle of incidence at face AB is",round(A1,0),"Degree"

(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]
(b) Angle of incidence at face AB is 73.0 Degree


## Example 21.43 Page no 660¶

In [38]:
#Given
u=40.0                         #cm
R=10.0
u1=1
u2=1.5

#Calculation
v=u2/(((u2-u1)/R)+(u1/u))

#Result
print"Position of the image is", v,"cm"

Position of the image is 20.0 cm


## Example 21.44 Page no 660¶

In [53]:
#Given
x=0.04
y=8/300.0
u=-0.04
v=-16/500.0

#Calculation
U=x/y
R=(1-U)/(-U/u+1/v)
f=1/((U-1)/R)

print"(i) Refractive index is",U
print"(iii) Focal length is",f,"m"

(i) Refractive index is 1.5
(ii) Radius of curvature is -0.08 m
(iii) Focal length is -0.16 m


## Example 21.46 Page no 661¶

In [71]:
#Given
OI=90.0                     #cm
O1O2=20
x=35                          #cm

#Calculation
f=(OI-x)*x/((OI-x)+x)

#Result
print"Focal length of the lens is", round(f,1),"cm"

Focal length of the lens is 21.4 cm


## Example 21.47 Page no 662¶

In [61]:
#Given
m=-4
a=1.5
u=0.3

#Calculation
x=a/(-m+1)
v=a-u
f=1/(1/u+1/v)

#Result
print"Focal length of the lens is", f,"m"

Focal length of the lens is 0.24 m


## Example 21.48 Page no 662¶

In [66]:
#Given
P1=8.0                        #dioptre
P2=2.0

#Calculation
f1=1/P1
f2=1/P2

#Result
print"Focal length of lens 1 is",f1,"m"
print"Focal length of lens 2 is",f2,"m"

Focal length of lens 1 is 0.125 m
Focal length of lens 2 is 0.5 m