Lens formula is the relation between the focal length of the lens with the distance of object and image.
For finding lens formula, we generally consider thin lenses because in case of thick lenses, it is not possible to find the exact distances of object, image and focal length and also the optical centre of the lens. Before deriving relation, we must know the sign conversions.
Sign Convention of lens :
1. All distances are measured from the optical centre of the lens.
2. The object is always placed on the left side of lens so that the direction of incident light is always from the left to right.
3. All distances measured from the optical centre of the lens to the right side are taken as positive.
4. All distances measured from the optical centre of the lens to the left side of lens are taken as negative.
5. Transverse upward distance from the principal axis is taken as the + ve and downward distance is taken as – ve.
Assumptions :
1. The lens is supposed to be thin.
2. Optical centre of the lens is taken at origin.
3. Incident rays are close to the principal axis of the lens.
4. All rays passing through optical centre of the lens go undeviated.
Lens Formula for Thin Convex Lens when real image is formed:
1/f = 1/v – 1/u
Lens Formula for Convex Lens when virtual image is formed :
1/v – 1/u = 1/f
Lens Formula for Thin Concave Lens :
1/v – 1/u = 1/f
Hence we conclude that lens formula for all types of lenses will be same i.e.,
1/v – 1/u = 1/f
irrespect time of the position of object and the size and nature of the image formed.
Practice Problem :
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, Calculate the object and image distance.
Solution: As the image formed is erect, it is virtual in nature. Therefore, both u and v are negative and positive and hence m is positive i.e., m = +4
Now v/u = m = 4
v = 4u
Now using lens formula 1/f = 1/v – 1/u
As f = 20 cm
so 1/20 = 1/4u – 1/u
= (1–4) / 4u = – 3/4u
1/20 = –3/4u
u = – 15 cm
v =4u
= 4 X (–15)
= –60 cm (Ans.)
Practice Questions:
1. An object is placed at focus of a concave mirror of focal length 20 cm. the distance of image is
(a) 40 cm (b) 10 cm
(c) 20 cm (d) infinity
2. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of the lens is :
(a) 30 cm, concave (b) 60 cm, concave
(c) 60 cm, convex (d) 30 cm, convex
3. The distance between an object and a divergent lens is 'm' times the focal length of the lens. the linear magnification produced by the lens is equal to
(a) m (b) 1/m
(c) (m+1) (d) 1/(m+1)
4. If magnification is negative, the nature of the image formed by a convex lens is
(a) real and inverted (b) virtual and erect
(c) real and erect (d) virtual and inverted
5. If magnification is positive, the nature of the image formed in convex lens is
(a) real and inverted (b) virtual and erect
(c) real (d) None of these
6. Magnification produced by a plane mirror is
(a) +1 (b) +2
(c) +1/2 (d) –1
7. What is the magnification when the object is placed at 2f from the pole of a concave mirror.
(a) 1/3 (b) 2/3
(c) 1 (d) 3/2
8. A lens which can produce magnification of 1, less than 1 or greater than 1 can be a
(a) convex lens (b) glass slab
(c) concave lens (d) plane mirror
9. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of glass is
(a) concave, 30 cm (b) convex, 30 cm
(c) concave, 20 cm (d) convex, 15 cm
10. The focal length of a convex lens depends upon
(a) wave length of light ray
(b) frequency of light ray
(c) time period of light ray
(d) velocity of light ray
11. As an object is brought closer to the focal point of a convex lens from infinity, its image
(a) becomes smaller (b) remains same
(c) gets farther from lens (d) gets closer to the lens
12. as an object is moved closer to the focal point of a convex lens from infinity, the size o image will
(a) decrease (b) increase
(c) become zero (d) none of the above
13. On decreasing the radii of the two surfaces of a double convex or double concave lens result in the decrease in the
(a) focal length (b) size of image
(c) size of object (d) intensity of image
14. Which of the following is wrong ? Here m represents the magnification of a lens
(a) The m is +ve when image formed by convex lens is virtual
(b) The m is –ve when image formed by concave lens is real
(c) The m is +ve when image is formed by concave lens
(d) None of the above
Answers:
1. d
2. d
3. d
4. a
5. b
6. a
7. c
8. a
9. b
10. a
11. c
12. b
13. a
14. d
Courtesy:R.A.Banwat
For finding lens formula, we generally consider thin lenses because in case of thick lenses, it is not possible to find the exact distances of object, image and focal length and also the optical centre of the lens. Before deriving relation, we must know the sign conversions.
Sign Convention of lens :
1. All distances are measured from the optical centre of the lens.
2. The object is always placed on the left side of lens so that the direction of incident light is always from the left to right.
3. All distances measured from the optical centre of the lens to the right side are taken as positive.
4. All distances measured from the optical centre of the lens to the left side of lens are taken as negative.
5. Transverse upward distance from the principal axis is taken as the + ve and downward distance is taken as – ve.
Assumptions :
1. The lens is supposed to be thin.
2. Optical centre of the lens is taken at origin.
3. Incident rays are close to the principal axis of the lens.
4. All rays passing through optical centre of the lens go undeviated.
Lens Formula for Thin Convex Lens when real image is formed:
1/f = 1/v – 1/u
Lens Formula for Convex Lens when virtual image is formed :
1/v – 1/u = 1/f
Lens Formula for Thin Concave Lens :
1/v – 1/u = 1/f
Hence we conclude that lens formula for all types of lenses will be same i.e.,
1/v – 1/u = 1/f
irrespect time of the position of object and the size and nature of the image formed.
Practice Problem :
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, Calculate the object and image distance.
Solution: As the image formed is erect, it is virtual in nature. Therefore, both u and v are negative and positive and hence m is positive i.e., m = +4
Now v/u = m = 4
v = 4u
Now using lens formula 1/f = 1/v – 1/u
As f = 20 cm
so 1/20 = 1/4u – 1/u
= (1–4) / 4u = – 3/4u
1/20 = –3/4u
u = – 15 cm
v =4u
= 4 X (–15)
= –60 cm (Ans.)
Practice Questions:
1. An object is placed at focus of a concave mirror of focal length 20 cm. the distance of image is
(a) 40 cm (b) 10 cm
(c) 20 cm (d) infinity
2. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of the lens is :
(a) 30 cm, concave (b) 60 cm, concave
(c) 60 cm, convex (d) 30 cm, convex
3. The distance between an object and a divergent lens is 'm' times the focal length of the lens. the linear magnification produced by the lens is equal to
(a) m (b) 1/m
(c) (m+1) (d) 1/(m+1)
4. If magnification is negative, the nature of the image formed by a convex lens is
(a) real and inverted (b) virtual and erect
(c) real and erect (d) virtual and inverted
5. If magnification is positive, the nature of the image formed in convex lens is
(a) real and inverted (b) virtual and erect
(c) real (d) None of these
6. Magnification produced by a plane mirror is
(a) +1 (b) +2
(c) +1/2 (d) –1
7. What is the magnification when the object is placed at 2f from the pole of a concave mirror.
(a) 1/3 (b) 2/3
(c) 1 (d) 3/2
8. A lens which can produce magnification of 1, less than 1 or greater than 1 can be a
(a) convex lens (b) glass slab
(c) concave lens (d) plane mirror
9. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of glass is
(a) concave, 30 cm (b) convex, 30 cm
(c) concave, 20 cm (d) convex, 15 cm
10. The focal length of a convex lens depends upon
(a) wave length of light ray
(b) frequency of light ray
(c) time period of light ray
(d) velocity of light ray
11. As an object is brought closer to the focal point of a convex lens from infinity, its image
(a) becomes smaller (b) remains same
(c) gets farther from lens (d) gets closer to the lens
12. as an object is moved closer to the focal point of a convex lens from infinity, the size o image will
(a) decrease (b) increase
(c) become zero (d) none of the above
13. On decreasing the radii of the two surfaces of a double convex or double concave lens result in the decrease in the
(a) focal length (b) size of image
(c) size of object (d) intensity of image
14. Which of the following is wrong ? Here m represents the magnification of a lens
(a) The m is +ve when image formed by convex lens is virtual
(b) The m is –ve when image formed by concave lens is real
(c) The m is +ve when image is formed by concave lens
(d) None of the above
Answers:
1. d
2. d
3. d
4. a
5. b
6. a
7. c
8. a
9. b
10. a
11. c
12. b
13. a
14. d
Courtesy:R.A.Banwat
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