NEET Exam  >  NEET Notes  >  DC Pandey Solutions for NEET Physics  >  DC Pandey Solutions: Refraction of Light- 2

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

Download, print and study this document offline
Please wait while the PDF view is loading
 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
Read More
122 docs

Up next

122 docs
Download as PDF

Up next

Explore Courses for NEET exam

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

practice quizzes

,

Previous Year Questions with Solutions

,

Extra Questions

,

Objective type Questions

,

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

,

past year papers

,

Important questions

,

Viva Questions

,

video lectures

,

ppt

,

Semester Notes

,

study material

,

mock tests for examination

,

Exam

,

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

,

MCQs

,

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

,

Summary

,

Free

,

shortcuts and tricks

,

pdf

,

Sample Paper

;