DC Pandey Solutions: Refraction of Light- 2 | DC Pandey Solutions for NEET Physics PDF Download

 Page 1


28. n
c
v
ice
2.3
1.30 = =
´
´
=
3 10
10
8
8
   q
c
n
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
1 1
ice
1.30
        =
-
sin ( )
1
0.77
29. (a) Let an gle of re frac tion in ma te rial 2 is r
then 
sin
sin
q
r
= =
1.8
1.6
18
16
 …(i)
For (2) to (3) interface
sin
sin
r
90
13
18 °
= =
1.3
1.8
Þ sin r =
13
18
 …(ii)
From (i) and (ii) sin q = ´
18
16
13
18
Þ    q =
æ
è
ç
ö
ø
÷
-
sin
1
13
16
(b) Yes, if q decreases r also decreases and
become less than the critical angle and
hence light goes into material 3.
30. Let maximum height of liq uid is h. From
figure for crit i cal an gle C
m = =
+
1
2 2
sin C
r h
r
 …(i)
Here r = 1 cm and m =
4
3
 putting these
values in Eq. (i). Solving we get h =
4
3
cm
Here sin q
m
c
g
= =
1 2
3
Now if water film is poured on the glass air
surface. Let emergent angle at glass water
surface is r, then
sin
sin
q m
m
c w
g
r
= =
´
´
4 2
3 3
Þ sin sin r
c
= = ´ =
9
8
9
8
2
3
3
4
q
Þ r =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
32. For to tal in ter nal re flec tion at top sur face
sin ( )
sin
90
90
1
2
° -
°
=
r n
n
Þ cos r
n
n
=
1
2
 and 
sin
sin
q
r
n
n
=
2
1
Þ sin sin q =
n
n
r
2
1
Þ      sin cos q = -
n
n
r
2
1
2
1
 = -
æ
è
ç
ç
ö
ø
÷
÷
n
n
n
n
2
1
1
2
2
1
Þ       sin q =
- n n
n
2
2
1
2
1
2
Þ             q =
æ
è
ç
ç
ö
ø
÷
÷
-
-
sin
1 2
1
2
1
n
n
33. The de vi a tion an gle vary from 0° to q° where 
q = ° - 90 c …(i)
where C is the critical angle
Now, sin
/
/
c
w
g
= = =
m
m
4 3
3 2
8
9
From Eq. (i) cos sin q = C
Þ cos q =
8
9
28
90–r
r
n
2
n
1
n
1
q
h
c
h
22
r + h
Page 2


28. n
c
v
ice
2.3
1.30 = =
´
´
=
3 10
10
8
8
   q
c
n
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
1 1
ice
1.30
        =
-
sin ( )
1
0.77
29. (a) Let an gle of re frac tion in ma te rial 2 is r
then 
sin
sin
q
r
= =
1.8
1.6
18
16
 …(i)
For (2) to (3) interface
sin
sin
r
90
13
18 °
= =
1.3
1.8
Þ sin r =
13
18
 …(ii)
From (i) and (ii) sin q = ´
18
16
13
18
Þ    q =
æ
è
ç
ö
ø
÷
-
sin
1
13
16
(b) Yes, if q decreases r also decreases and
become less than the critical angle and
hence light goes into material 3.
30. Let maximum height of liq uid is h. From
figure for crit i cal an gle C
m = =
+
1
2 2
sin C
r h
r
 …(i)
Here r = 1 cm and m =
4
3
 putting these
values in Eq. (i). Solving we get h =
4
3
cm
Here sin q
m
c
g
= =
1 2
3
Now if water film is poured on the glass air
surface. Let emergent angle at glass water
surface is r, then
sin
sin
q m
m
c w
g
r
= =
´
´
4 2
3 3
Þ sin sin r
c
= = ´ =
9
8
9
8
2
3
3
4
q
Þ r =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
32. For to tal in ter nal re flec tion at top sur face
sin ( )
sin
90
90
1
2
° -
°
=
r n
n
Þ cos r
n
n
=
1
2
 and 
sin
sin
q
r
n
n
=
2
1
Þ sin sin q =
n
n
r
2
1
Þ      sin cos q = -
n
n
r
2
1
2
1
 = -
æ
è
ç
ç
ö
ø
÷
÷
n
n
n
n
2
1
1
2
2
1
Þ       sin q =
- n n
n
2
2
1
2
1
2
Þ             q =
æ
è
ç
ç
ö
ø
÷
÷
-
-
sin
1 2
1
2
1
n
n
33. The de vi a tion an gle vary from 0° to q° where 
q = ° - 90 c …(i)
where C is the critical angle
Now, sin
/
/
c
w
g
= = =
m
m
4 3
3 2
8
9
From Eq. (i) cos sin q = C
Þ cos q =
8
9
28
90–r
r
n
2
n
1
n
1
q
h
c
h
22
r + h
Þ q =
æ
è
ç
ö
ø
÷
-
cos
1
8
9
Hence deviation angle vary from 0° to 
cos .
- æ
è
ç
ö
ø
÷
1
8
9
34.
(a)Only circular patch light escapes
because only those rays which are
incident within a cone of semivertex
angle C [Critical angle] are refracted out
of the water surface. All other rays are
totally internally reflected as shown in
figures
(b) Now m = =
+
1
2 2
sin C
r h
r
or     C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
2 2
r
r h
35.
For maximum angle q the angle 90 - r at left
surface must be equal to critical angle
Þ sin ( ) 90
1 100
125
4
5
° - = = = r
1.25
Þ cos r =
4
5
Þ sin r =
3
5
Now, 
sin
sin
q
r
= = 1.25
5
4
Þ sin sin q = = ´
5
4
5
4
3
5
r 
Þ q =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
36. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A m
A
2
2
3
2
2
2
2 2
2
=
+ æ
è
ç
ö
ø
÷
=
sin
sin
sin
sin cos
sin
A
A
A A
A
Þ cos
A
2
3
2
= Þ 
A
2
30 = °
Þ A = ° 60
37. Here i r
1 1
0 = = ° .
Now, let other face angle of incidence is r
2
Q r r A
1 2
+ =  Þ 0
2
+ = r A
Þ       r A
2
30 = = °
sin
sin
r
i
2
2
1
=
1.5
 Þ sin sin i r
2 2
= 1.4
Þ sin sin i
2
30 = ´ ° 1.5
Þ       i
2
1
19 = = °
-
sin (0.75)
38. From figure sin Ð = Ð OQP OQR
Hence the ray retrace its path.
39.
 29
45°
r
1
R
P
r
1 45°
r
2
r
1
45°
Q
i
2
90°–r
r
90°–r
q
90
m = 1.0
m = 1.25
h
S
r
h
c
c
30°
60°
30°
90° R
B
A
Q
45°
Page 3


28. n
c
v
ice
2.3
1.30 = =
´
´
=
3 10
10
8
8
   q
c
n
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
1 1
ice
1.30
        =
-
sin ( )
1
0.77
29. (a) Let an gle of re frac tion in ma te rial 2 is r
then 
sin
sin
q
r
= =
1.8
1.6
18
16
 …(i)
For (2) to (3) interface
sin
sin
r
90
13
18 °
= =
1.3
1.8
Þ sin r =
13
18
 …(ii)
From (i) and (ii) sin q = ´
18
16
13
18
Þ    q =
æ
è
ç
ö
ø
÷
-
sin
1
13
16
(b) Yes, if q decreases r also decreases and
become less than the critical angle and
hence light goes into material 3.
30. Let maximum height of liq uid is h. From
figure for crit i cal an gle C
m = =
+
1
2 2
sin C
r h
r
 …(i)
Here r = 1 cm and m =
4
3
 putting these
values in Eq. (i). Solving we get h =
4
3
cm
Here sin q
m
c
g
= =
1 2
3
Now if water film is poured on the glass air
surface. Let emergent angle at glass water
surface is r, then
sin
sin
q m
m
c w
g
r
= =
´
´
4 2
3 3
Þ sin sin r
c
= = ´ =
9
8
9
8
2
3
3
4
q
Þ r =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
32. For to tal in ter nal re flec tion at top sur face
sin ( )
sin
90
90
1
2
° -
°
=
r n
n
Þ cos r
n
n
=
1
2
 and 
sin
sin
q
r
n
n
=
2
1
Þ sin sin q =
n
n
r
2
1
Þ      sin cos q = -
n
n
r
2
1
2
1
 = -
æ
è
ç
ç
ö
ø
÷
÷
n
n
n
n
2
1
1
2
2
1
Þ       sin q =
- n n
n
2
2
1
2
1
2
Þ             q =
æ
è
ç
ç
ö
ø
÷
÷
-
-
sin
1 2
1
2
1
n
n
33. The de vi a tion an gle vary from 0° to q° where 
q = ° - 90 c …(i)
where C is the critical angle
Now, sin
/
/
c
w
g
= = =
m
m
4 3
3 2
8
9
From Eq. (i) cos sin q = C
Þ cos q =
8
9
28
90–r
r
n
2
n
1
n
1
q
h
c
h
22
r + h
Þ q =
æ
è
ç
ö
ø
÷
-
cos
1
8
9
Hence deviation angle vary from 0° to 
cos .
- æ
è
ç
ö
ø
÷
1
8
9
34.
(a)Only circular patch light escapes
because only those rays which are
incident within a cone of semivertex
angle C [Critical angle] are refracted out
of the water surface. All other rays are
totally internally reflected as shown in
figures
(b) Now m = =
+
1
2 2
sin C
r h
r
or     C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
2 2
r
r h
35.
For maximum angle q the angle 90 - r at left
surface must be equal to critical angle
Þ sin ( ) 90
1 100
125
4
5
° - = = = r
1.25
Þ cos r =
4
5
Þ sin r =
3
5
Now, 
sin
sin
q
r
= = 1.25
5
4
Þ sin sin q = = ´
5
4
5
4
3
5
r 
Þ q =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
36. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A m
A
2
2
3
2
2
2
2 2
2
=
+ æ
è
ç
ö
ø
÷
=
sin
sin
sin
sin cos
sin
A
A
A A
A
Þ cos
A
2
3
2
= Þ 
A
2
30 = °
Þ A = ° 60
37. Here i r
1 1
0 = = ° .
Now, let other face angle of incidence is r
2
Q r r A
1 2
+ =  Þ 0
2
+ = r A
Þ       r A
2
30 = = °
sin
sin
r
i
2
2
1
=
1.5
 Þ sin sin i r
2 2
= 1.4
Þ sin sin i
2
30 = ´ ° 1.5
Þ       i
2
1
19 = = °
-
sin (0.75)
38. From figure sin Ð = Ð OQP OQR
Hence the ray retrace its path.
39.
 29
45°
r
1
R
P
r
1 45°
r
2
r
1
45°
Q
i
2
90°–r
r
90°–r
q
90
m = 1.0
m = 1.25
h
S
r
h
c
c
30°
60°
30°
90° R
B
A
Q
45°
The ray re trace its path from ref. by sur face 
AB hence Ð = ° ARq 90 from ge om e try it is
clear that r = ° 30
m =
sin
sin
i
r
Þ m =
°
°
sin
sin
45
30
 Þ m =
1 2
1 2
/
/
 Þ m = 2
40. Depends on formula.
41. The maximum an gle will be A C = 2 where C
is the critical angle
Now, C =
æ
è
ç
ö
ø
÷
=
-
sin
1
1
1.5
41.81
Hence A C = = ´ = 2 2 41.81 83.62
42. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A
A
m
2
2
 here A = ° 60
1.5 =
° +
æ
è
ç
ö
ø
÷
°
sin
sin
60
2
30
d
m
0.75 =
° +
æ
è
ç
ö
ø
÷ sin
60
2
d
m
Þ 60 2
1
° + =
-
d
m
sin ( ) 0.75
Þ d
m
= 22.8
and not deviation = ° - = ° 180 22.8 157.2
(b) If the system is placed in water
m = =
1.5
4/3
4.5
4
Þ 60 2 30
1
° + ¢ = ´ °
-
d
m
sin ( sin ) 1.125
Þ d
m
¢ =
é
ë
ê
ù
û
ú
- °
-
2 60
1
sin
1.125
2
Net deviation = - = ° 180 d
m
128.4
43. w
m m
m
=
-
-
V R
Y
1
0.0305
1.665 1.645
=
-
- m
y
1
On solving we get my = 1.656
44.
w w
1
1
2
2
0
f f
+ =
Þ 
0.18
20 30
2
+
-
w
 Þ w
2
30
20
=
´ 0.18
Þ w
2
= 0.27
Now, 
1 1 1
1 2
F f f
= + Þ 
1 1
20
1
30 F
= -
Þ F = 60 cm
45.
w w
1
1
2
2
0
f f
+ =
Þ  
w
w
1
2
1
2
= -
f
f
    
3
2
1
2
= -
f
f
Þ     f f
1 2
3
2
= -
Now,  
1 1 1
1 2
F f f
= +
Þ   
1
150
1 2
3
2 2
= -
f f
f
2
50 = cm and f
1
75 = - cm
46. Applying m =
sin
sin
i
r
1
1
Find angle r
1
 for two different refraction
indices. Because i
1
65 = ° from both the
cases.
Then again apply
m =
sin
sin
i
r
2
2
 and find i
2
. Because r A r
2 1
= -
.
Then apply :
d = + - ( ) i i A
1 2
for two refraction indices. Then difference
in deviations is :
Dd d d = -
1 2
30
Page 4


28. n
c
v
ice
2.3
1.30 = =
´
´
=
3 10
10
8
8
   q
c
n
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
1 1
ice
1.30
        =
-
sin ( )
1
0.77
29. (a) Let an gle of re frac tion in ma te rial 2 is r
then 
sin
sin
q
r
= =
1.8
1.6
18
16
 …(i)
For (2) to (3) interface
sin
sin
r
90
13
18 °
= =
1.3
1.8
Þ sin r =
13
18
 …(ii)
From (i) and (ii) sin q = ´
18
16
13
18
Þ    q =
æ
è
ç
ö
ø
÷
-
sin
1
13
16
(b) Yes, if q decreases r also decreases and
become less than the critical angle and
hence light goes into material 3.
30. Let maximum height of liq uid is h. From
figure for crit i cal an gle C
m = =
+
1
2 2
sin C
r h
r
 …(i)
Here r = 1 cm and m =
4
3
 putting these
values in Eq. (i). Solving we get h =
4
3
cm
Here sin q
m
c
g
= =
1 2
3
Now if water film is poured on the glass air
surface. Let emergent angle at glass water
surface is r, then
sin
sin
q m
m
c w
g
r
= =
´
´
4 2
3 3
Þ sin sin r
c
= = ´ =
9
8
9
8
2
3
3
4
q
Þ r =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
32. For to tal in ter nal re flec tion at top sur face
sin ( )
sin
90
90
1
2
° -
°
=
r n
n
Þ cos r
n
n
=
1
2
 and 
sin
sin
q
r
n
n
=
2
1
Þ sin sin q =
n
n
r
2
1
Þ      sin cos q = -
n
n
r
2
1
2
1
 = -
æ
è
ç
ç
ö
ø
÷
÷
n
n
n
n
2
1
1
2
2
1
Þ       sin q =
- n n
n
2
2
1
2
1
2
Þ             q =
æ
è
ç
ç
ö
ø
÷
÷
-
-
sin
1 2
1
2
1
n
n
33. The de vi a tion an gle vary from 0° to q° where 
q = ° - 90 c …(i)
where C is the critical angle
Now, sin
/
/
c
w
g
= = =
m
m
4 3
3 2
8
9
From Eq. (i) cos sin q = C
Þ cos q =
8
9
28
90–r
r
n
2
n
1
n
1
q
h
c
h
22
r + h
Þ q =
æ
è
ç
ö
ø
÷
-
cos
1
8
9
Hence deviation angle vary from 0° to 
cos .
- æ
è
ç
ö
ø
÷
1
8
9
34.
(a)Only circular patch light escapes
because only those rays which are
incident within a cone of semivertex
angle C [Critical angle] are refracted out
of the water surface. All other rays are
totally internally reflected as shown in
figures
(b) Now m = =
+
1
2 2
sin C
r h
r
or     C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
2 2
r
r h
35.
For maximum angle q the angle 90 - r at left
surface must be equal to critical angle
Þ sin ( ) 90
1 100
125
4
5
° - = = = r
1.25
Þ cos r =
4
5
Þ sin r =
3
5
Now, 
sin
sin
q
r
= = 1.25
5
4
Þ sin sin q = = ´
5
4
5
4
3
5
r 
Þ q =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
36. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A m
A
2
2
3
2
2
2
2 2
2
=
+ æ
è
ç
ö
ø
÷
=
sin
sin
sin
sin cos
sin
A
A
A A
A
Þ cos
A
2
3
2
= Þ 
A
2
30 = °
Þ A = ° 60
37. Here i r
1 1
0 = = ° .
Now, let other face angle of incidence is r
2
Q r r A
1 2
+ =  Þ 0
2
+ = r A
Þ       r A
2
30 = = °
sin
sin
r
i
2
2
1
=
1.5
 Þ sin sin i r
2 2
= 1.4
Þ sin sin i
2
30 = ´ ° 1.5
Þ       i
2
1
19 = = °
-
sin (0.75)
38. From figure sin Ð = Ð OQP OQR
Hence the ray retrace its path.
39.
 29
45°
r
1
R
P
r
1 45°
r
2
r
1
45°
Q
i
2
90°–r
r
90°–r
q
90
m = 1.0
m = 1.25
h
S
r
h
c
c
30°
60°
30°
90° R
B
A
Q
45°
The ray re trace its path from ref. by sur face 
AB hence Ð = ° ARq 90 from ge om e try it is
clear that r = ° 30
m =
sin
sin
i
r
Þ m =
°
°
sin
sin
45
30
 Þ m =
1 2
1 2
/
/
 Þ m = 2
40. Depends on formula.
41. The maximum an gle will be A C = 2 where C
is the critical angle
Now, C =
æ
è
ç
ö
ø
÷
=
-
sin
1
1
1.5
41.81
Hence A C = = ´ = 2 2 41.81 83.62
42. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A
A
m
2
2
 here A = ° 60
1.5 =
° +
æ
è
ç
ö
ø
÷
°
sin
sin
60
2
30
d
m
0.75 =
° +
æ
è
ç
ö
ø
÷ sin
60
2
d
m
Þ 60 2
1
° + =
-
d
m
sin ( ) 0.75
Þ d
m
= 22.8
and not deviation = ° - = ° 180 22.8 157.2
(b) If the system is placed in water
m = =
1.5
4/3
4.5
4
Þ 60 2 30
1
° + ¢ = ´ °
-
d
m
sin ( sin ) 1.125
Þ d
m
¢ =
é
ë
ê
ù
û
ú
- °
-
2 60
1
sin
1.125
2
Net deviation = - = ° 180 d
m
128.4
43. w
m m
m
=
-
-
V R
Y
1
0.0305
1.665 1.645
=
-
- m
y
1
On solving we get my = 1.656
44.
w w
1
1
2
2
0
f f
+ =
Þ 
0.18
20 30
2
+
-
w
 Þ w
2
30
20
=
´ 0.18
Þ w
2
= 0.27
Now, 
1 1 1
1 2
F f f
= + Þ 
1 1
20
1
30 F
= -
Þ F = 60 cm
45.
w w
1
1
2
2
0
f f
+ =
Þ  
w
w
1
2
1
2
= -
f
f
    
3
2
1
2
= -
f
f
Þ     f f
1 2
3
2
= -
Now,  
1 1 1
1 2
F f f
= +
Þ   
1
150
1 2
3
2 2
= -
f f
f
2
50 = cm and f
1
75 = - cm
46. Applying m =
sin
sin
i
r
1
1
Find angle r
1
 for two different refraction
indices. Because i
1
65 = ° from both the
cases.
Then again apply
m =
sin
sin
i
r
2
2
 and find i
2
. Because r A r
2 1
= -
.
Then apply :
d = + - ( ) i i A
1 2
for two refraction indices. Then difference
in deviations is :
Dd d d = -
1 2
30
Objective Questions (Level-1)
1. Endoscope is bases on to tal in ter nal
refraction Hence, correct option is (c)
2. Here m
l
= + A
B
2
Q m is dimensionless.
Þ 
B
l
2
= dimension of m
Þ B = l
2
 Þ B has dimension of Area
Hence, correct option is (d).
3. Shift = -
æ
è
ç
ç
ö
ø
÷
÷
1
1
m
Q m
R
 is minimum. than other visible colour. 
Red colour least raised.
correct option is (c)
4. Critical an gle q
m
C
=
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
Q m
0
 is maximum for violet colour hence q
c
for violet colour is least.
Hence correct option is (d)
5. We have P
f
=
1
(metre)
= = ´ - +
é
ë
ê
ù
û
ú
100
100 1
1 1
f
n
R r ( )
( )
cm
Þ P =
´ ´
= +
100 2
10
12
0.6
Hence, correct option is (a).
6. Speed of light in wa ter =
c
w
m
Þ v
w
=
´
= ´
3 10
4 3
10
8
8
/
2.25 m/s
Hence correct option os (c).
7. Due to TIR emer gent beam will turn into
black.
Hence correct option is (c).
8. Q v n = l but fre quency n re main constant and 
v de creases hence l de creases.
Hence correct option is (b).
9. Using Snell’s law
 On first and 2nd  interface
sin
sin
i
q
m
m
=
2
1
 …(i)
and 
sin
sin
q m
m r
=
3
2
 …(ii)
Multiplying (i) and (ii), we get
sin
sin
i
r
=
m
m
3
1
Hence correct option is (b).
10. We have i r =
1
 and r i
2
90 = -
Now 
sin
sin
i
r
2
1
=
m
Þ 
sin
sin ( )
i
i 90
1
° -
=
m
Þ tan i =
1
m
 …(i)
If C is the critical angle then C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
 
Þ C i =
-
sin (tan )
1
Hence correct option is (a).
 31
q
q
m
2
r
m
3
m
1
i
1 2
90°
i
r
2
B A
C
r
1
Denser medium
Rarer medium
Page 5


28. n
c
v
ice
2.3
1.30 = =
´
´
=
3 10
10
8
8
   q
c
n
=
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
1 1
ice
1.30
        =
-
sin ( )
1
0.77
29. (a) Let an gle of re frac tion in ma te rial 2 is r
then 
sin
sin
q
r
= =
1.8
1.6
18
16
 …(i)
For (2) to (3) interface
sin
sin
r
90
13
18 °
= =
1.3
1.8
Þ sin r =
13
18
 …(ii)
From (i) and (ii) sin q = ´
18
16
13
18
Þ    q =
æ
è
ç
ö
ø
÷
-
sin
1
13
16
(b) Yes, if q decreases r also decreases and
become less than the critical angle and
hence light goes into material 3.
30. Let maximum height of liq uid is h. From
figure for crit i cal an gle C
m = =
+
1
2 2
sin C
r h
r
 …(i)
Here r = 1 cm and m =
4
3
 putting these
values in Eq. (i). Solving we get h =
4
3
cm
Here sin q
m
c
g
= =
1 2
3
Now if water film is poured on the glass air
surface. Let emergent angle at glass water
surface is r, then
sin
sin
q m
m
c w
g
r
= =
´
´
4 2
3 3
Þ sin sin r
c
= = ´ =
9
8
9
8
2
3
3
4
q
Þ r =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
32. For to tal in ter nal re flec tion at top sur face
sin ( )
sin
90
90
1
2
° -
°
=
r n
n
Þ cos r
n
n
=
1
2
 and 
sin
sin
q
r
n
n
=
2
1
Þ sin sin q =
n
n
r
2
1
Þ      sin cos q = -
n
n
r
2
1
2
1
 = -
æ
è
ç
ç
ö
ø
÷
÷
n
n
n
n
2
1
1
2
2
1
Þ       sin q =
- n n
n
2
2
1
2
1
2
Þ             q =
æ
è
ç
ç
ö
ø
÷
÷
-
-
sin
1 2
1
2
1
n
n
33. The de vi a tion an gle vary from 0° to q° where 
q = ° - 90 c …(i)
where C is the critical angle
Now, sin
/
/
c
w
g
= = =
m
m
4 3
3 2
8
9
From Eq. (i) cos sin q = C
Þ cos q =
8
9
28
90–r
r
n
2
n
1
n
1
q
h
c
h
22
r + h
Þ q =
æ
è
ç
ö
ø
÷
-
cos
1
8
9
Hence deviation angle vary from 0° to 
cos .
- æ
è
ç
ö
ø
÷
1
8
9
34.
(a)Only circular patch light escapes
because only those rays which are
incident within a cone of semivertex
angle C [Critical angle] are refracted out
of the water surface. All other rays are
totally internally reflected as shown in
figures
(b) Now m = =
+
1
2 2
sin C
r h
r
or     C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
2 2
r
r h
35.
For maximum angle q the angle 90 - r at left
surface must be equal to critical angle
Þ sin ( ) 90
1 100
125
4
5
° - = = = r
1.25
Þ cos r =
4
5
Þ sin r =
3
5
Now, 
sin
sin
q
r
= = 1.25
5
4
Þ sin sin q = = ´
5
4
5
4
3
5
r 
Þ q =
æ
è
ç
ö
ø
÷
-
sin
1
3
4
36. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A m
A
2
2
3
2
2
2
2 2
2
=
+ æ
è
ç
ö
ø
÷
=
sin
sin
sin
sin cos
sin
A
A
A A
A
Þ cos
A
2
3
2
= Þ 
A
2
30 = °
Þ A = ° 60
37. Here i r
1 1
0 = = ° .
Now, let other face angle of incidence is r
2
Q r r A
1 2
+ =  Þ 0
2
+ = r A
Þ       r A
2
30 = = °
sin
sin
r
i
2
2
1
=
1.5
 Þ sin sin i r
2 2
= 1.4
Þ sin sin i
2
30 = ´ ° 1.5
Þ       i
2
1
19 = = °
-
sin (0.75)
38. From figure sin Ð = Ð OQP OQR
Hence the ray retrace its path.
39.
 29
45°
r
1
R
P
r
1 45°
r
2
r
1
45°
Q
i
2
90°–r
r
90°–r
q
90
m = 1.0
m = 1.25
h
S
r
h
c
c
30°
60°
30°
90° R
B
A
Q
45°
The ray re trace its path from ref. by sur face 
AB hence Ð = ° ARq 90 from ge om e try it is
clear that r = ° 30
m =
sin
sin
i
r
Þ m =
°
°
sin
sin
45
30
 Þ m =
1 2
1 2
/
/
 Þ m = 2
40. Depends on formula.
41. The maximum an gle will be A C = 2 where C
is the critical angle
Now, C =
æ
è
ç
ö
ø
÷
=
-
sin
1
1
1.5
41.81
Hence A C = = ´ = 2 2 41.81 83.62
42. m
d
=
+ æ
è
ç
ö
ø
÷ sin
sin
A
A
m
2
2
 here A = ° 60
1.5 =
° +
æ
è
ç
ö
ø
÷
°
sin
sin
60
2
30
d
m
0.75 =
° +
æ
è
ç
ö
ø
÷ sin
60
2
d
m
Þ 60 2
1
° + =
-
d
m
sin ( ) 0.75
Þ d
m
= 22.8
and not deviation = ° - = ° 180 22.8 157.2
(b) If the system is placed in water
m = =
1.5
4/3
4.5
4
Þ 60 2 30
1
° + ¢ = ´ °
-
d
m
sin ( sin ) 1.125
Þ d
m
¢ =
é
ë
ê
ù
û
ú
- °
-
2 60
1
sin
1.125
2
Net deviation = - = ° 180 d
m
128.4
43. w
m m
m
=
-
-
V R
Y
1
0.0305
1.665 1.645
=
-
- m
y
1
On solving we get my = 1.656
44.
w w
1
1
2
2
0
f f
+ =
Þ 
0.18
20 30
2
+
-
w
 Þ w
2
30
20
=
´ 0.18
Þ w
2
= 0.27
Now, 
1 1 1
1 2
F f f
= + Þ 
1 1
20
1
30 F
= -
Þ F = 60 cm
45.
w w
1
1
2
2
0
f f
+ =
Þ  
w
w
1
2
1
2
= -
f
f
    
3
2
1
2
= -
f
f
Þ     f f
1 2
3
2
= -
Now,  
1 1 1
1 2
F f f
= +
Þ   
1
150
1 2
3
2 2
= -
f f
f
2
50 = cm and f
1
75 = - cm
46. Applying m =
sin
sin
i
r
1
1
Find angle r
1
 for two different refraction
indices. Because i
1
65 = ° from both the
cases.
Then again apply
m =
sin
sin
i
r
2
2
 and find i
2
. Because r A r
2 1
= -
.
Then apply :
d = + - ( ) i i A
1 2
for two refraction indices. Then difference
in deviations is :
Dd d d = -
1 2
30
Objective Questions (Level-1)
1. Endoscope is bases on to tal in ter nal
refraction Hence, correct option is (c)
2. Here m
l
= + A
B
2
Q m is dimensionless.
Þ 
B
l
2
= dimension of m
Þ B = l
2
 Þ B has dimension of Area
Hence, correct option is (d).
3. Shift = -
æ
è
ç
ç
ö
ø
÷
÷
1
1
m
Q m
R
 is minimum. than other visible colour. 
Red colour least raised.
correct option is (c)
4. Critical an gle q
m
C
=
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
Q m
0
 is maximum for violet colour hence q
c
for violet colour is least.
Hence correct option is (d)
5. We have P
f
=
1
(metre)
= = ´ - +
é
ë
ê
ù
û
ú
100
100 1
1 1
f
n
R r ( )
( )
cm
Þ P =
´ ´
= +
100 2
10
12
0.6
Hence, correct option is (a).
6. Speed of light in wa ter =
c
w
m
Þ v
w
=
´
= ´
3 10
4 3
10
8
8
/
2.25 m/s
Hence correct option os (c).
7. Due to TIR emer gent beam will turn into
black.
Hence correct option is (c).
8. Q v n = l but fre quency n re main constant and 
v de creases hence l de creases.
Hence correct option is (b).
9. Using Snell’s law
 On first and 2nd  interface
sin
sin
i
q
m
m
=
2
1
 …(i)
and 
sin
sin
q m
m r
=
3
2
 …(ii)
Multiplying (i) and (ii), we get
sin
sin
i
r
=
m
m
3
1
Hence correct option is (b).
10. We have i r =
1
 and r i
2
90 = -
Now 
sin
sin
i
r
2
1
=
m
Þ 
sin
sin ( )
i
i 90
1
° -
=
m
Þ tan i =
1
m
 …(i)
If C is the critical angle then C =
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
1
m
 
Þ C i =
-
sin (tan )
1
Hence correct option is (a).
 31
q
q
m
2
r
m
3
m
1
i
1 2
90°
i
r
2
B A
C
r
1
Denser medium
Rarer medium
11. Let an gle of min i mum deviation is d
m
we know that
m
d
=
+ æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
sin
sin
A
A
m
2
2
Þ 2
60
2
30
=
° +
°
sin
( )
sin
d
m
Þ 
1
2
60
2
=
° +
æ
è
ç
ö
ø
÷ sin
d
m
 
Þ 
60
2
45
+
= °
d
m
 Þ d
m
= ° 30
Hence correct option is (a).
12. We know that
   
1
1
1 1
1 2
f
n
R R
= - -
æ
è
ç
ç
ö
ø
÷
÷
( )
       
1
15 1
1 1
2
1 2
a R R
= - -
æ
è
ç
ç
ö
ø
÷
÷
( . )
Þ 
1
0 1
1 1
1 2
0.2
.5 = - -
æ
è
ç
ç
ö
ø
÷
÷
( )
R R
 ...(i)
Let refractive index of the liquid s n
l
 
Þ 
1
1
1 1
1 2
f
n
n R R
cm l
= -
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
ç
ö
ø
÷
÷
Þ - = -
æ
è
ç
ç
ö
ø
÷
÷
´
1
1 10
0.5
1.5
n
l
Þ   
-
= -
1
5
1
1.5
n
l
Þ   
1.5 
n
l
=
4
5
 Þ 
5
4
´
=
1.5
n
l
  Þ  n
l
=
15
8
Hence correct option is (b).
13.
We have, 
sin
sin
i
r
1
1
=
m
m
 …(i)
sin
sin
r
r
1
2
2
1
=
m
m
 …(ii)
and 
sin
sin
r
r
2
3
3
2
=
m
m
 …(iii)
sin
sin
r
x
3 4
3
=
m
m
 …(iv)
Multiplying (i), (ii), (iii) and (iv), we get
sin
sin
i
x
=
m
m
4
  Þ  sin sin x i =
m
m
4
Hence correct option is (b)
14. Let ra dius of cur va ture of the lens is R
then 
1
1
1 1
f
n
R R
= - +
æ
è
ç
ö
ø
÷
( ) Þ f
R
x
=
- 2 1 ( )
Let focal length of one part is f ¢
then 
1
1
1 1
f
n
R ¢
= - -
¥
é
ë
ê
ù
û
ú
( )
Þ f
R
n
f ¢ =
-
= ¢
( ) 1
2
The focal length of the combination is
1 1
2
1
2
1
2
1
2 F f f f f
= + + + Þ F
f
=
2
Hence correct option is (b).
15. Here P = + 5D Þ f + = 20 cm
Þ     
1
20
1
1 1
1 2
= - -
é
ë
ê
ù
û
ú
( ) 1.5
R R
…(i)
and - = -
æ
è
ç
ç
ö
ø
÷
÷
-
é
ë
ê
ù
û
ú
1
100
1
1 1
1 2
1.5
n R R
e
 …(ii)
Dividing Eq. (ii) by (i) - =
-
æ
è
ç
ç
ö
ø
÷
÷
1
5
1
1.5
0.5
n
l
Þ          
1
10
1 + =
1.5
n
l
Þ   n
l
=
5
3
Hence correct option is (b).
32
m
1
m
i
r
1
r
r
2
m
2
m
3
x
x
r
3
m
4