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JEE Solved Example Based on Ray Optics 
JEE Mains 
Q1: Two plane mirrors are inclined at ????
°
. A ray incident on one mirror at an angle of ?? after 
reflection falls on the second mirror and is reflected from there parallel to the first mirror, ?? is: 
(A) ????
°
 
(B) ????
°
 
(C) ????
°
 
(D) ????
°
 
Ans: (A) 
?? 1
= 180
°
- 20; ?? 2
= 180
°
- 20
'
= 70
°
+ 90
°
- ?? '
 
 
 ? ?? '
= 20
°
?? = 180
°
- 70
°
- ( 90
°
- ?? '
)= 180
°
- 70
°
- 70
°
?? = 40
°
? ?? = 90
°
- ?? = 50
°
 
 
 
Q2:  There are two plane mirror with reflecting surfaces facing each other. Both the mirrors are 
moving with the speed of ?? away from each other. A point object is placed between the mirrors. 
The velocity of the image formed due to the ?? th 
 reflection will be 
(A) ???? 
(B) ?? ???? 
(C) ?? ???? 
(D) ?? ???? 
Page 2


JEE Solved Example Based on Ray Optics 
JEE Mains 
Q1: Two plane mirrors are inclined at ????
°
. A ray incident on one mirror at an angle of ?? after 
reflection falls on the second mirror and is reflected from there parallel to the first mirror, ?? is: 
(A) ????
°
 
(B) ????
°
 
(C) ????
°
 
(D) ????
°
 
Ans: (A) 
?? 1
= 180
°
- 20; ?? 2
= 180
°
- 20
'
= 70
°
+ 90
°
- ?? '
 
 
 ? ?? '
= 20
°
?? = 180
°
- 70
°
- ( 90
°
- ?? '
)= 180
°
- 70
°
- 70
°
?? = 40
°
? ?? = 90
°
- ?? = 50
°
 
 
 
Q2:  There are two plane mirror with reflecting surfaces facing each other. Both the mirrors are 
moving with the speed of ?? away from each other. A point object is placed between the mirrors. 
The velocity of the image formed due to the ?? th 
 reflection will be 
(A) ???? 
(B) ?? ???? 
(C) ?? ???? 
(D) ?? ???? 
 
 
Ans:  (B) With respect to mirror1 the object is going array from mirror. So first image will also more 
away w.r.t mirror 1 with same speed ?? . So with respect to object ?? the image speed is 
 
?? image1 
= ?? image1,mirror1 
+ ?? mirror 1,obj 
= ?? + ?? = 2?? 
Now this image becomes object for mirror 2. With respect to mirror 2 the image is going away 
towards 
 
 
Q3: A man of height ' ?? ' is walking away from a street lamp with a constant speed ' ?? '. The height 
of the street lamp is ?? ?? . The rate at which the length of the man's shadow is increasing when he is 
at a distance of ???? ?? from the base of the street lamp is: 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? ?? 
(D) ?? /?? 
Ans:  (A) tan ?? =
h
?? =
3h
10h+?? lim
?? ?8
? ? 10h + ?? = 3?? ? ?? = 5h 
For general case 
 
Page 3


JEE Solved Example Based on Ray Optics 
JEE Mains 
Q1: Two plane mirrors are inclined at ????
°
. A ray incident on one mirror at an angle of ?? after 
reflection falls on the second mirror and is reflected from there parallel to the first mirror, ?? is: 
(A) ????
°
 
(B) ????
°
 
(C) ????
°
 
(D) ????
°
 
Ans: (A) 
?? 1
= 180
°
- 20; ?? 2
= 180
°
- 20
'
= 70
°
+ 90
°
- ?? '
 
 
 ? ?? '
= 20
°
?? = 180
°
- 70
°
- ( 90
°
- ?? '
)= 180
°
- 70
°
- 70
°
?? = 40
°
? ?? = 90
°
- ?? = 50
°
 
 
 
Q2:  There are two plane mirror with reflecting surfaces facing each other. Both the mirrors are 
moving with the speed of ?? away from each other. A point object is placed between the mirrors. 
The velocity of the image formed due to the ?? th 
 reflection will be 
(A) ???? 
(B) ?? ???? 
(C) ?? ???? 
(D) ?? ???? 
 
 
Ans:  (B) With respect to mirror1 the object is going array from mirror. So first image will also more 
away w.r.t mirror 1 with same speed ?? . So with respect to object ?? the image speed is 
 
?? image1 
= ?? image1,mirror1 
+ ?? mirror 1,obj 
= ?? + ?? = 2?? 
Now this image becomes object for mirror 2. With respect to mirror 2 the image is going away 
towards 
 
 
Q3: A man of height ' ?? ' is walking away from a street lamp with a constant speed ' ?? '. The height 
of the street lamp is ?? ?? . The rate at which the length of the man's shadow is increasing when he is 
at a distance of ???? ?? from the base of the street lamp is: 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? ?? 
(D) ?? /?? 
Ans:  (A) tan ?? =
h
?? =
3h
10h+?? lim
?? ?8
? ? 10h + ?? = 3?? ? ?? = 5h 
For general case 
 
tan ?? =
3h
?? + ?? =
h
?? ? 3?? = ?? + ?? ? 2?? = ?? ? 2 ·
????
????
=
????
????
?
????
????
=
1
2
· ?? 
 
 
Q4: A boy of height ?? . ?? ?? with his eye level at ?? . ?? ?? stands before a plane mirror of length 
?? . ???? ?? fixed on the wall. The height of the lower edge of the mirror above the floor is ?? . ?? ?? . 
Then, 
(A) The boy will see his full image. 
(B) The boy cannot see his hair. 
(C) The boy cannot see his feet. 
(D) The boy cannot see neither his hair nor his feet. 
Ans: (C) A is head and ?? is feet of man. ?? is the eye. The mirror can be placed anywhere between the 
centre line ???? (of ???? ) and ???? (of ???? ) to get full image from head to feet. 
So here ???? = 1.4 m. So ???? should be 0.7 m. But mirror is 0.8 m from ground so feet will not be 
visible. The upper edge of mirror is at height ( 0.8 + 0.75) m equal to 1.55 m which is more than ???? . 
 
???? = ???? + ???? =
????
2
+ ???? =
0.1
2
+ 1.4
 = 0.05 + 1.4 = 1.45 m
 
So head will be visible. 
 
 
 
Page 4


JEE Solved Example Based on Ray Optics 
JEE Mains 
Q1: Two plane mirrors are inclined at ????
°
. A ray incident on one mirror at an angle of ?? after 
reflection falls on the second mirror and is reflected from there parallel to the first mirror, ?? is: 
(A) ????
°
 
(B) ????
°
 
(C) ????
°
 
(D) ????
°
 
Ans: (A) 
?? 1
= 180
°
- 20; ?? 2
= 180
°
- 20
'
= 70
°
+ 90
°
- ?? '
 
 
 ? ?? '
= 20
°
?? = 180
°
- 70
°
- ( 90
°
- ?? '
)= 180
°
- 70
°
- 70
°
?? = 40
°
? ?? = 90
°
- ?? = 50
°
 
 
 
Q2:  There are two plane mirror with reflecting surfaces facing each other. Both the mirrors are 
moving with the speed of ?? away from each other. A point object is placed between the mirrors. 
The velocity of the image formed due to the ?? th 
 reflection will be 
(A) ???? 
(B) ?? ???? 
(C) ?? ???? 
(D) ?? ???? 
 
 
Ans:  (B) With respect to mirror1 the object is going array from mirror. So first image will also more 
away w.r.t mirror 1 with same speed ?? . So with respect to object ?? the image speed is 
 
?? image1 
= ?? image1,mirror1 
+ ?? mirror 1,obj 
= ?? + ?? = 2?? 
Now this image becomes object for mirror 2. With respect to mirror 2 the image is going away 
towards 
 
 
Q3: A man of height ' ?? ' is walking away from a street lamp with a constant speed ' ?? '. The height 
of the street lamp is ?? ?? . The rate at which the length of the man's shadow is increasing when he is 
at a distance of ???? ?? from the base of the street lamp is: 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? ?? 
(D) ?? /?? 
Ans:  (A) tan ?? =
h
?? =
3h
10h+?? lim
?? ?8
? ? 10h + ?? = 3?? ? ?? = 5h 
For general case 
 
tan ?? =
3h
?? + ?? =
h
?? ? 3?? = ?? + ?? ? 2?? = ?? ? 2 ·
????
????
=
????
????
?
????
????
=
1
2
· ?? 
 
 
Q4: A boy of height ?? . ?? ?? with his eye level at ?? . ?? ?? stands before a plane mirror of length 
?? . ???? ?? fixed on the wall. The height of the lower edge of the mirror above the floor is ?? . ?? ?? . 
Then, 
(A) The boy will see his full image. 
(B) The boy cannot see his hair. 
(C) The boy cannot see his feet. 
(D) The boy cannot see neither his hair nor his feet. 
Ans: (C) A is head and ?? is feet of man. ?? is the eye. The mirror can be placed anywhere between the 
centre line ???? (of ???? ) and ???? (of ???? ) to get full image from head to feet. 
So here ???? = 1.4 m. So ???? should be 0.7 m. But mirror is 0.8 m from ground so feet will not be 
visible. The upper edge of mirror is at height ( 0.8 + 0.75) m equal to 1.55 m which is more than ???? . 
 
???? = ???? + ???? =
????
2
+ ???? =
0.1
2
+ 1.4
 = 0.05 + 1.4 = 1.45 m
 
So head will be visible. 
 
 
 
Q5:  A point source of light ?? is placed in front of two large mirrors as shown. Which of the 
following observers will see only one image of ?? ? 
(A) Only A 
(B) Only ?? 
(C) Both A and C 
(B) Both B and C 
 
Ans: (B) Rays from S going right from the normal will not reach the bottom horizontal mirror as they 
will hit the inclined mirror and get deviated. So C will see only the image formed by inclined mirror. 
 
 
 
Q6:  In the figure shown if the object ' ?? ' moves toward the plane mirror, then the image ?? (which 
is formed after successive reflections from ?? ?? &?? ?? , respectively) will move: 
(A) Toward right 
(B) Toward left 
(C) With zero velocity 
(D) Cannot be determined 
 
Ans:  (A) Object O moves towards ?? 1
 so image 1 due to ?? 1
 will move towards left i.e. towards ?? 2
. 
For ?? 2
 we have formula for speed of image as 
????
????
= - (
?? 2
?? 2
)
????
????
. So negative sign means final image 2 
due to ?? 2
 will move opposite to image 1 i.e. towards right. 
 
 
Page 5


JEE Solved Example Based on Ray Optics 
JEE Mains 
Q1: Two plane mirrors are inclined at ????
°
. A ray incident on one mirror at an angle of ?? after 
reflection falls on the second mirror and is reflected from there parallel to the first mirror, ?? is: 
(A) ????
°
 
(B) ????
°
 
(C) ????
°
 
(D) ????
°
 
Ans: (A) 
?? 1
= 180
°
- 20; ?? 2
= 180
°
- 20
'
= 70
°
+ 90
°
- ?? '
 
 
 ? ?? '
= 20
°
?? = 180
°
- 70
°
- ( 90
°
- ?? '
)= 180
°
- 70
°
- 70
°
?? = 40
°
? ?? = 90
°
- ?? = 50
°
 
 
 
Q2:  There are two plane mirror with reflecting surfaces facing each other. Both the mirrors are 
moving with the speed of ?? away from each other. A point object is placed between the mirrors. 
The velocity of the image formed due to the ?? th 
 reflection will be 
(A) ???? 
(B) ?? ???? 
(C) ?? ???? 
(D) ?? ???? 
 
 
Ans:  (B) With respect to mirror1 the object is going array from mirror. So first image will also more 
away w.r.t mirror 1 with same speed ?? . So with respect to object ?? the image speed is 
 
?? image1 
= ?? image1,mirror1 
+ ?? mirror 1,obj 
= ?? + ?? = 2?? 
Now this image becomes object for mirror 2. With respect to mirror 2 the image is going away 
towards 
 
 
Q3: A man of height ' ?? ' is walking away from a street lamp with a constant speed ' ?? '. The height 
of the street lamp is ?? ?? . The rate at which the length of the man's shadow is increasing when he is 
at a distance of ???? ?? from the base of the street lamp is: 
(A) ?? /?? 
(B) ?? /?? 
(C) ?? ?? 
(D) ?? /?? 
Ans:  (A) tan ?? =
h
?? =
3h
10h+?? lim
?? ?8
? ? 10h + ?? = 3?? ? ?? = 5h 
For general case 
 
tan ?? =
3h
?? + ?? =
h
?? ? 3?? = ?? + ?? ? 2?? = ?? ? 2 ·
????
????
=
????
????
?
????
????
=
1
2
· ?? 
 
 
Q4: A boy of height ?? . ?? ?? with his eye level at ?? . ?? ?? stands before a plane mirror of length 
?? . ???? ?? fixed on the wall. The height of the lower edge of the mirror above the floor is ?? . ?? ?? . 
Then, 
(A) The boy will see his full image. 
(B) The boy cannot see his hair. 
(C) The boy cannot see his feet. 
(D) The boy cannot see neither his hair nor his feet. 
Ans: (C) A is head and ?? is feet of man. ?? is the eye. The mirror can be placed anywhere between the 
centre line ???? (of ???? ) and ???? (of ???? ) to get full image from head to feet. 
So here ???? = 1.4 m. So ???? should be 0.7 m. But mirror is 0.8 m from ground so feet will not be 
visible. The upper edge of mirror is at height ( 0.8 + 0.75) m equal to 1.55 m which is more than ???? . 
 
???? = ???? + ???? =
????
2
+ ???? =
0.1
2
+ 1.4
 = 0.05 + 1.4 = 1.45 m
 
So head will be visible. 
 
 
 
Q5:  A point source of light ?? is placed in front of two large mirrors as shown. Which of the 
following observers will see only one image of ?? ? 
(A) Only A 
(B) Only ?? 
(C) Both A and C 
(B) Both B and C 
 
Ans: (B) Rays from S going right from the normal will not reach the bottom horizontal mirror as they 
will hit the inclined mirror and get deviated. So C will see only the image formed by inclined mirror. 
 
 
 
Q6:  In the figure shown if the object ' ?? ' moves toward the plane mirror, then the image ?? (which 
is formed after successive reflections from ?? ?? &?? ?? , respectively) will move: 
(A) Toward right 
(B) Toward left 
(C) With zero velocity 
(D) Cannot be determined 
 
Ans:  (A) Object O moves towards ?? 1
 so image 1 due to ?? 1
 will move towards left i.e. towards ?? 2
. 
For ?? 2
 we have formula for speed of image as 
????
????
= - (
?? 2
?? 2
)
????
????
. So negative sign means final image 2 
due to ?? 2
 will move opposite to image 1 i.e. towards right. 
 
 
Q7:  A point source of light is ???? ???? away from a screen and is kept at the focus of a concave 
mirror that reflects light on the screen. The focal length of the mirror is 20 ???? . The ratio of average 
intensities of the illumination on the screen when the mirror is present and when the mirror is 
removed is: 
(A) ???? : ?? 
(B) ???? : ?? 
(C) ???? : ?? 
(D) ???? : ?? 
Ans:  (D) Intensity incident 
=
P?? 2
4 × (area on which lightis incident )
 
When the mirror is not present, light is reaching the screen up to height h. Maximum area on which 
light is incident = ?? h
2
 
Intensity =
P?? 2
4?? h
2
 and 
tan ?? =
h
60
?
Ph
2
4?? h
2
× ( 60)
2
=
P
4?? × ( 60)
2
 
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FAQs on Ray Optics Solved Examples - Physics for JEE Main & Advanced

1. What are the basic principles of ray optics?
Ans. Ray optics is based on the principles of reflection, refraction, and the behavior of light rays as they pass through different mediums. These principles help in understanding how light behaves when interacting with objects and surfaces.
2. How is a ray diagram helpful in understanding ray optics?
Ans. A ray diagram is a visual representation of how light rays interact with objects such as mirrors and lenses. It helps in predicting the path of light rays, locating images, and understanding the behavior of light in different situations.
3. What is the difference between a real image and a virtual image in ray optics?
Ans. A real image is formed when light rays actually converge at a specific point, whereas a virtual image is formed when light rays appear to diverge from a point. Real images can be projected onto a screen, while virtual images cannot.
4. How does the focal length of a lens affect the formation of images in ray optics?
Ans. The focal length of a lens determines the point at which light rays converge or diverge after passing through the lens. A shorter focal length results in a more convergent lens, while a longer focal length results in a more divergent lens, affecting the size and location of images formed.
5. Can ray optics be applied to other electromagnetic waves besides visible light?
Ans. Yes, the principles of ray optics can be applied to other electromagnetic waves such as radio waves, microwaves, and X-rays. These waves also exhibit behaviors of reflection, refraction, and diffraction similar to visible light.
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