Ray Optics and Optical Instruments Practice Questions - DPP for JEE

 Page 1


1. (a) From mirror formula
 so, 
?
? m/s
2. (a) We have,
or  
or 
or 
or .
3. (b) f
0
 = 100 cm, f
e
 = 5 cm
When final image is formed at least distance of distinct vision (d), then
Page 2


1. (a) From mirror formula
 so, 
?
? m/s
2. (a) We have,
or  
or 
or 
or .
3. (b) f
0
 = 100 cm, f
e
 = 5 cm
When final image is formed at least distance of distinct vision (d), then
M = [ Q D = 25 cm]
M = 
4. (c) I
1
 is the image of object O formed by the lens.
;  u
1
 = – 15,  f
1
 = 10
Solving we get,  v
1
 = 30 cm.
I
1
 acts as source for mirror
?   u
2
 = – (45 – v
1
) = – 15 cm.
I
2
 is the image formed by the mirror
?  
? v
2
 = –30cm
The height of I
2
 above principal axis of lens is
= 
I
2
 acts as a source for lens, u
3
 = –(45 – v
2
) = –15 cm.
Hence, the lens forms an image I
3
 at a distance v
3
 = 30 cm to the left of
lens.
The height of I
3
 below  the principal axis of lens
Page 3


1. (a) From mirror formula
 so, 
?
? m/s
2. (a) We have,
or  
or 
or 
or .
3. (b) f
0
 = 100 cm, f
e
 = 5 cm
When final image is formed at least distance of distinct vision (d), then
M = [ Q D = 25 cm]
M = 
4. (c) I
1
 is the image of object O formed by the lens.
;  u
1
 = – 15,  f
1
 = 10
Solving we get,  v
1
 = 30 cm.
I
1
 acts as source for mirror
?   u
2
 = – (45 – v
1
) = – 15 cm.
I
2
 is the image formed by the mirror
?  
? v
2
 = –30cm
The height of I
2
 above principal axis of lens is
= 
I
2
 acts as a source for lens, u
3
 = –(45 – v
2
) = –15 cm.
Hence, the lens forms an image I
3
 at a distance v
3
 = 30 cm to the left of
lens.
The height of I
3
 below  the principal axis of lens
= 
?  Required distance = 
5. (d) The focal length of the lens
  
Shift = 
= 
Now v’ = 
Now the object be distance u'.
u' = –7 ×16 × 5 = – 560 cm = – 5.6 m
6. (c)
Applying Snell’s law at Q
Page 4


1. (a) From mirror formula
 so, 
?
? m/s
2. (a) We have,
or  
or 
or 
or .
3. (b) f
0
 = 100 cm, f
e
 = 5 cm
When final image is formed at least distance of distinct vision (d), then
M = [ Q D = 25 cm]
M = 
4. (c) I
1
 is the image of object O formed by the lens.
;  u
1
 = – 15,  f
1
 = 10
Solving we get,  v
1
 = 30 cm.
I
1
 acts as source for mirror
?   u
2
 = – (45 – v
1
) = – 15 cm.
I
2
 is the image formed by the mirror
?  
? v
2
 = –30cm
The height of I
2
 above principal axis of lens is
= 
I
2
 acts as a source for lens, u
3
 = –(45 – v
2
) = –15 cm.
Hence, the lens forms an image I
3
 at a distance v
3
 = 30 cm to the left of
lens.
The height of I
3
 below  the principal axis of lens
= 
?  Required distance = 
5. (d) The focal length of the lens
  
Shift = 
= 
Now v’ = 
Now the object be distance u'.
u' = –7 ×16 × 5 = – 560 cm = – 5.6 m
6. (c)
Applying Snell’s law at Q
  ...(1)
Applying Snell’s Law at P
 ; from (1)
or       
7. (b) From the fig.
Angle of deviation, 
Here,  e = i
and 
? 
For equilateral prism, A = 60°
? 
8. (a) For point A, 
For point B, sin (90° – r) = 
g
µ
a
?  
where,
(90° – r) is critical angle.
Page 5


1. (a) From mirror formula
 so, 
?
? m/s
2. (a) We have,
or  
or 
or 
or .
3. (b) f
0
 = 100 cm, f
e
 = 5 cm
When final image is formed at least distance of distinct vision (d), then
M = [ Q D = 25 cm]
M = 
4. (c) I
1
 is the image of object O formed by the lens.
;  u
1
 = – 15,  f
1
 = 10
Solving we get,  v
1
 = 30 cm.
I
1
 acts as source for mirror
?   u
2
 = – (45 – v
1
) = – 15 cm.
I
2
 is the image formed by the mirror
?  
? v
2
 = –30cm
The height of I
2
 above principal axis of lens is
= 
I
2
 acts as a source for lens, u
3
 = –(45 – v
2
) = –15 cm.
Hence, the lens forms an image I
3
 at a distance v
3
 = 30 cm to the left of
lens.
The height of I
3
 below  the principal axis of lens
= 
?  Required distance = 
5. (d) The focal length of the lens
  
Shift = 
= 
Now v’ = 
Now the object be distance u'.
u' = –7 ×16 × 5 = – 560 cm = – 5.6 m
6. (c)
Applying Snell’s law at Q
  ...(1)
Applying Snell’s Law at P
 ; from (1)
or       
7. (b) From the fig.
Angle of deviation, 
Here,  e = i
and 
? 
For equilateral prism, A = 60°
? 
8. (a) For point A, 
For point B, sin (90° – r) = 
g
µ
a
?  
where,
(90° – r) is critical angle.
? 
 
= 
 
    
9. (b) u = –50 cm = –0.5 m
v = –30 cm = –0.3 m
 .
10. (b) For the image of point P to be seen by the observer, it should be
formed at point Q.
In ?QNS,
NS = QS = 2h