Contents
- 1 What is Longitudinal Magnification? What Type of Graph do you Expect Between 1 U and 1V?
- 1.1 Areal Magnification of the Image of an Area Kept Perpendicular to the Principal Axis
- 1.2 Longitudinal or Axial Magnification of the Image of an Object Kept Along the Principal Axis
- 1.3 Relation of f and y or f and u with m
- 1.4 u ~ v and \(\frac{1}{u}\) ~ \(\frac{1}{v}\) Graphs for Convex Lens
- 1.5 Rules for solving numerical problems of lens:
- 1.6 Numerical Examples
- 1.7 Image of a Virtual Object
- 1.8 Numerical Examples
Physics Topics can help us understand the behavior of the natural world around us.
What is Longitudinal Magnification? What Type of Graph do you Expect Between 1 U and 1V?
Linear or Transverse or Lateral Magnification of the Image of an Object Kept Perpendicular to the Principal Axis
Definition: Linear magnification of an image formed by a lens is defined as the ratio of the size of the image to the size of the object.
Denoting linear magnification by m we have, from Fig.’s,
m \(=\frac{\text { size of image }(I)}{\text { size of object }(O)}\) = \(\frac{p q}{P Q}\) = \(\frac{v}{u}\) \(=\frac{\text { image distance }}{\text { object distance }}\)
According to sign convention :
- For formation of real image in convex lens u is negative and v is positive [Fig], So linear magnification m is negative. The image is inverted.
- For formation of virtual image in convex lens both n and v are negative [Fig.]. So linear magnification m is positive. The image is erect.
- In case of concave lens both n and v are negative, [Fig.]. So linear magnification m is positive. So the image is erect. So we can say that if magnification is negative, the image is inverted and if magnification is positive, the image is erect.
We can determine the expression for magnification of image formed by a lens by the same process as adopted for determination of the magnification of image formed by reflection of light in curved surface. It is to be noted that the real image formed by reflection is formed in front of the mirror i.e. on the same side of the mirror as the object. But in case of lens, the real image formed by a lens due to refraction is formed on the opposite side of the real object.
The general expression for magnification of the image formed by refraction in lens is given by m = \(\frac{I}{O}\) = \(\frac{v}{u}\)
Magnification produced by a combination of lenses: The magnification of the final image produced by a combination of lenses is given by m = m1 × m2 × m3 ; where m1, m2, m3,… etc. are respectively the magnifications produced by each lens.
Areal Magnification of the Image of an Area Kept Perpendicular to the Principal Axis
Definition: Linear magnification of an image formed by a lens is defined as the ratio of the size of the image to the size of the object.
Let the length and breadth of a two dimensional object be l and b respectively. Hence, the area of the object, A = lb .
If the linear magnification of the image be m, length of the image, l’ = m × l and breadth of the image, b’ = m × b
∴ Area of the image, A’ = l’b’ = m2lb = m2A
Therefore areal magnification,
m’ = \(\frac{A^{\prime}}{A}\) = m2 …. (1)
Longitudinal or Axial Magnification of the Image of an Object Kept Along the Principal Axis
Definition: Longitudinal or axial magnification of the image formed by a lens of an object kept along the principal axis is defined as the ratio of the length of image to that of the object.
Let an extended object is kept along the principal axis of a convex lens.
Let u1 and u2 be the distances of the nearest and the furthest points respectively of the object along the principal axis and v1 and v2 the respective image distances [Fig.].
∴ Longitudinal magnification, m” = \(\frac{v_2-v_1}{u_2-u_1}\) = \(\frac{\Delta v}{\Delta u}\)
For infinitesimal values of Δv and ΔM , magnification should be noted as \(\frac{d v}{d u}\).
Differentiating the lens equation, \(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{f}\), with respect to u we get,
\(-\frac{1}{v^2} \frac{d v}{d u}-\frac{1}{u^2}\) = 0 [∵ f is constant]
or, \(\frac{d v}{d u}\) = \(\frac{-v^2}{u^2}\) ∴ m” = \(\frac{d v}{d u}\) = -m2 …. (1)
∴ Longitudinal magnification = -(linear magnification)2
Hence, longitudinal magnification in case of lens is numerically equal to the square of linear magnification.
i) It is clear from equation (1) that m” is always negative irrespective of the sign of m . This implies that object and image always lie in the opposite directions along the principal axis, whatever may be the nature of the image—real or virtual, in convex or concave lens. This matter is called axial change.
Relation of f and y or f and u with m
The lens formula is
u ~ v and \(\frac{1}{u}\) ~ \(\frac{1}{v}\) Graphs for Convex Lens
The focal length of convex lens f is the positive. The real image formed by this lens is always situated on the side opposite to the object. So image distance v is also positive. According to the sign convention object distance u is negative. Hence if a real image is formed by a convex lens the equation of the lens is as follows:
\(\frac{1}{v}\) – \(\frac{1}{-u}\) = \(\frac{1}{f}\) or, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
u ~ v graph: In case of convex lens if for different values of object distances and the corresponding image distances are recorded and plotted on a graph, it will be a rectangular hyperbola [Fig.].
\(\frac{1}{u}\) ~ \(\frac{1}{v}\) graph: In case of covex lens if \(\frac{1}{u}\) ~ \(\frac{1}{v}\) graph is drawn taking different values of u and v, it will be a straight line [Fig.]. The intercept cut by the straight line AB from the axes are each equal to \(\frac{1}{f}\).
i.e., OA = OB = \(\frac{1}{f}\)
Rules for solving numerical problems of lens:
1. The numerical values of u, v, f with their proper signs should be put in the general equation of lens. The sign of the quantity which has to be determined, need no sign.
2. After solving the equation, conclusion is to be taken about the sign of the solved quantity.
(a) If the sign of f is positive, the lens will be convex and if it is negative, the lens will be concave.
(b) If v is positive, the image will be virtual and erect. Again if v is negative, the image will be real and inverted.
3. To find the relation of u and v from the given value of m (magnification), then only the numerical value is to be taken.
4. If sign of m is positive, the image will be erect and if it is negative, the image will be inverted.
Numerical Examples
Example 1.
The distance of an object from a convex lens is 20 cm. If the focal length of the lens is 15 cm, determine the position of the image and its nature. [HS ’05]
Solution:
Here, u = -20 cm; as the lens is convex, f = +15 cm
we know that, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{1}{v}\) = \(\frac{1}{u}\) + \(\frac{1}{f}\) = \(\frac{1}{-20}+\frac{1}{15}\) = \(\frac{4-3}{60}\) = \(\frac{1}{60}\) or, v = 60 cm
As v is positive, the image will be formed on the side opposite to the object at a distance of 60 cm i.e., the image is real.
Magnification, m = \(\frac{1}{u}\) = \(\frac{60}{-20}\) = -3
So, an image magnified three times as the size of the object is formed. As magnification is negative, the image is inverted.
Example 2.
If an object is placed at a distance of 30 cm from a lens, a virtual image is formed. If the magnification of the image is \(\frac{2}{3}\), find the position of the image and the focal length of the lens. Also find the nature of the lens.
Solution:
Here, object distance, u = -30 cm, magnification m = \(\frac{m}{u}\) = \(\frac{2}{3}\) = \(\frac{v}{30}\) or, v = 20 cm
We know, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{u}{f}\) – 1 = \(\frac{u}{f}\) or, \(\frac{1}{m}\) – 1 = \(\frac{u}{f}\)
Substituting m = \(\frac{2}{3}\) and u = -30 we get,
\(\frac{3}{2}\) – 1 = \(\frac{-30}{f}\) or, f = -30 × 2 = -60 cm
∴ Focal length of the lens is 60 cm .
Further the negative sign of f implies that the lens is a concave one.
Example 3.
A convex lens forms a real image of an object magnified n times. Prove that the object distance = (n + 1)\(\frac{f}{n}\); f = focal length of the lens.
Solution:
Here, magnification = n
i.e., \(\frac{v}{u}\) = n or, v = nu
For a real object, u is negative and v is positive. For convex lens f is positive. Following this sign convention, we get from the lens formula,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{n u}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1+n}{n u}\) = \(\frac{1}{f}\)
or, u = (n + 1)\(\frac{f}{n}\)
i.e., the object distance is (n + 1)\(\frac{f}{n}\)
Example 4.
(i) A luminous object and a screen are placed 90 cm apart. To cast an image magnified twice the size of the object on the screen, what type of lens is required and what will be its focal length? [NCERT; CBSE ’10]
(ii) An object is situated at a distance of 10 m from a convex lens. A magnified image is to be cast on a screen by the lens. If magnification is 19, what is the focal length of the lens ? [NCERT]
Solution:
i) Since the image is formed on a screen, it is real. Hence the lens to be used should be convex.
In this case, u + v = 90 cm
and \(\frac{v}{u}\) = 2 or, v = 2u
∴ 3u = 90 cm or, u = 30 cm
∴ v = 90 – 30 = 60 cm
Substituting in lens formula, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
or \(\frac{1}{60}\) – \(\frac{1}{-30}\) = \(\frac{1}{f}\) [u = -30, as the object is real]
∴ Focal length of the lens is 20 cm
ii) Magnification, m = \(\frac{v}{u}\) = 19
∴ v = 19u = 19 × 10 = 190 m
Substituting v = 190 and u = -10 in \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
we get, \(\frac{1}{190}\) + \(\frac{1}{10}\) = \(\frac{1}{f}\) or, f = \(\frac{190}{1+19}\) = 9.5 m
∴ Focal length of the lens is 9.5 m.
Example 5.
A convex lens is placed just above an empty vessel. An object is placed at the bottom of the vessel and at a distance of 45 cm below the lens. An image of the object is formed above the vessel at a distance of 36 cm from the lens. A liquid is poured up to a height of 40 cm in the vessel. Now the image is formed above the vessel at a distance of 48 cm from the lens. Calculate the refractive index of the liquid.
Solution:
When the vessel is empty, u = (-45) cm, v = 36 cm
From the equation of the lens we get,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{1}{+36}-\frac{1}{-45}\) = \(\frac{1}{f}\) or, \(\frac{+9}{180}\) = \(\frac{1}{f}\) or, f = +20 cm
On pouring the liquid into the vessel the apparent position of the object will be raised.
Real depth of the liquid = 40 cm and apparent depth = x cm (say)
If refractive index of the liquid is µ, then
µ \(=\frac{\text { real depth }}{\text { apparent depth }}\) = \(\frac{40}{x}\) or, x = \(\frac{40}{\mu}\)
Now, distance of the lens from the liquid surface = 45 – 40 = 5 cm
∴ Object distance from the lens = (5 + x) = 5 + \(\frac{40}{\mu}\)
In this case, v = +48 cm and f = +20 cm
From the equation of the lens,
Example 6.
Two convex lenses of focal lengths 15 cm and 10 cm are placed coaxially. A ray of light parallel to the principal axis of a lens is incident on it and emerges from the other lens parallel to the same axis. Draw a neat ray-diagram. What is the distance of separation between the lenses?
Solution:
The ray incident on the lens L1 passes through the second principal focus of this lens after refraction. Since the ray after refraction through the second lens moves parallel to the principal axis of this lens, F is the first principal focus of lens L2 [Fig.].
So, O1F = 15 cm, O2F = 10 cm
∴ Distance of separation between the lenses = 15 + 10 = 25 cm
Example 7.
If an object is placed at a distance of 20 cm in front of a convex lens, a three times magnified and inverted image is formed. In which direction and how far is the lens to be moved to obtain an erect image of equal magnification [m = 3] ?
Solution:
Here, m = 3 and u = 20 cm
∴ \(\frac{v}{u}\) = 3 or, v = 3u = 3 × 20 = 60 cm
Now, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{60}\) – \(\frac{1}{-20}\) = \(\frac{1}{f}\) [following sign convention, u = -20 cm and v = 60 cm ]
f = 15 cm
To obtain an erect image, the lens is to be moved towards the object, so that the object distance now becomes less than the focal distance of the lens.
Let the lens is moved x cm towards the object.
So, the object distance becomes, u1 = (20 – x) cm .
Let the image distance be v1 cm.
∴ \(\frac{v_1}{u_1}\) = m = 3 or, v1 = 3u1 = 3(20 – x) cm
From lens equation following sign convention we have,
\(\frac{1}{v_1}-\frac{1}{u_1}\) = \(\frac{1}{f}\) or, \(\frac{1}{-3(20-x)}-\frac{1}{-(20-x)}\) = \(\frac{1}{15}\)
[as the image is virtual and so u1 is taken as negative]
or, \(\frac{2}{3(20-x)}\) = \(\frac{1}{15}\)
or, x = 10 cm
∴ The lens is to be moved by a distance 10 cm towards the lens.
Example 8.
If a magnifying lens of focal length 10 cm is held infront of very small writing, it is magnified five times. How far was the magnifying lens held?
Solution:
Here, focal length, f = 10 cm
Let object distance = u
Now, magnification, m = \(\frac{v}{u}\) = 5, or, u = 5u
Now, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{-5 u}-\frac{1}{-u}\) = \(\frac{1}{10}\)
[for magnifying lens, u and 11 both are negative]
or, \(-\frac{1}{5 u}+\frac{1}{u}\) = \(\frac{1}{10}\) or, \(\frac{4}{5 u}\) = \(\frac{1}{10}\) or, u = 8 cm
So, the lens was held at a distance of 8 cm from the writing.
Example 9.
An object is placed at a distance of 150 cm from a screen. A convex lens is placed between the object and the screen so that an image magnified 4 times the object is formed on the screen. Determine the position of the lens and Its focal length. [NCERT]
Solution:
Here, u + u = 150 cm
and magnification, m = 4 or, \(\frac{v}{u}\) = 4 or, u = 4u
From equation (1) we have,
u + 4u = 150
or, 5u = 150 or, u = 30 cm
∴ v = 4u = 4 × 30 = 120 cm
Hence, the distance of the lens from the screen = 120 cm
Now, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or \(\frac{1}{f}\) = \(\frac{1+4}{120}\) = \(\frac{5}{120}\) = \(\frac{5}{120}\)
or, f = 24 cm
Since the object and the image are situated on mutually opposite sides of the lens, hence u = -30 cm, u = 120 cm
or, \(\frac{1}{120}+\frac{1}{30}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{1+4}{120}\) = \(\frac{5}{120}\) = \(\frac{5}{120}\)
or, f = 24 cm
∴ Focal length of the lens is 24 cm.
Example 10.
A convex lens of focal length f forms an image which is m times magnified on a screen. If the distance of the object and the screen is x, prove that, f = \(\frac{m x}{(1+m)^2}\) [HS(XI) ‘09, HS ‘04]
Solution:
Since the image Is formed on a screen, it is real. So the image distance is positive. The focal length is also positive. Object distance is negative.
Magnification, m = \(\frac{v}{u}\) or, v = mu
Here, u + v = x or, u + mu = x or, u = \(\frac{x}{1+m}\)
∴ v = \(\frac{m x}{1+m}\)
The equation of lens : \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{m u}+\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1+m}{m u}\) = \(\frac{1}{f}\)
or, f = \(\frac{m u}{1+m}\) = \(\frac{m x}{(1+m)^2}\) [∵ u = \(\frac{x}{1+m}\)]
Example 11.
An object is placed on the left side of a convex lens. A of focal length 20 cm at a distance of 10 cm from the lens. Another convex lens of focal length 10 cm is placed co-axially on the right side of the lens A at a distance of 5 cm from It. Determine magnification and position of the final Image by the lens-combination.
Solution:
In case of image formation by the first lens, u = -10 cm; f = 20 cm, v = ?
Now, \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{v}-\frac{1}{-10}\) = \(+\frac{1}{20}\) or, \(\frac{1}{v}\) = \(\frac{-1}{10}+\frac{1}{20}\) = \(\frac{-1}{20}\) or, v = -20 cm
Since v is negative, a virtual image would form on the same side of the object at a distance of 20 cm from the lens.
This image will act as the object for the second lens.
In case of image formation by the second lens,
u = -(20 + 5) = -25 cm; f = 10 cm; v = ?
From the equation of lens,
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{v}-\frac{1}{-25}\) = \(\frac{1}{10}\) or, \(\frac{1}{v}\) = \(\frac{-1}{25}+\frac{1}{10}\) = \(\frac{3}{50}\)
or, v = \(\frac{50}{3}\) = 16.66 cm
So, finally a real image is formed on the right side of the second lens (on the opposite side of the object) at a distance of 16.66 cm from this lens.
Magnification by the first lens,
m1 = \(\frac{v}{u}\) = \(\frac{20}{10}\) = 2
And magnification by the second lens,
m2 = \(\frac{v}{u}\) = \(\frac{50}{3 \times 25}\) = \(\frac{2}{3}\)
∴ Magnification by the lens combination,
m = m1 × m2 = 2 × \(\frac{2}{3}\) = \(\frac{4}{3}\) = 1.33
Example 12.
A convex lens forms a real image of an object magnified 10 times. If the focal length of the lens is 20 cm, determine the distance of the object from the lens.
Solution:
Let object distance = X
Here, m = 10
∴ \(\frac{v}{x}\) = 10 or, v = 10x
In this case, v is positive and f = 20 cm
From the equation of the lens,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{10 x}-\frac{1}{-x}\) = \(\frac{1}{20}\)
or, \(\frac{11}{10 x}\) = \(\frac{1}{20}\)
∴ Object distance = 22 cm
Example 13.
Two convex lenses of focal lengths 3 cm and 4 cm are placed 8 cm apart from each other. An object of height 1 cm is placed in front of a lens of smaller focal length at a distance 4 cm. Determine magnification and size of the final image by the lens combination.
Solution:
In case of image formation by the first lens, u = -4 cm, f = 3 cm
Now, \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{1}{v}+\frac{1}{4}\) = \(\frac{1}{3}\) or \(\frac{1}{v}\) = \(\frac{1}{3}-\frac{1}{4}\) = \(\frac{1}{12}\)
or v = 12 cm
Since v is positive, an image would form on the side of the first lens opposite that of the object. This image acts as virtual image for second lens.
In case of image formation by this lens,
u = (12 – 8) = 4 cm, f = 4 cm, v = ?
From the equation of lens,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{v}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\) or, \(\frac{1}{v}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{1}{2}\)
So, final image is real and it will be formed at a distance 2cm from the second lens.
Magnification by the first lens,
m1 = \(\frac{v}{u}\) = \(\frac{12}{4}\) = 3
Magnification by the second lens,
m2 = \(\frac{v}{u}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
So, magnification by the lens combination,
m = m1 × m2 = 3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)
Again, m \(=\frac{\text { size of image }}{\text { size of object }}\)
or, \(\frac{3}{2}\) \(=\frac{\text { size of image }}{1}\)
or, size of image = \(\frac{3}{2}\) = 1.5 cm
Image of a Virtual Object
We have so far discussed image formation of real objects. But objects may be virtual as well and images of the virtual objects can be formed by using lenses.
In Fig.(a) and (b), a converging beam of rays is incident on a convex lens and a concave lens respectively. In absence of the lenses, the converging beam of rays would meet at Q on the other side of the lenses but due to the presence of the lenses the beam meets at Q’. So, Q is virtual object here and its image Q’ is real.
After refraction in the convex lens a convergent beam of rays becomes more convergent i.e., the convergent beam meets nearer to the lens. In case of the convex lens, Q’ is nearer from the lens than Q. Obviously in this case image distance, OQ’ is less than object distance, OQ. In convex lens the real image of a virtual object is formed and the image lies within the focus.
After refraction in the concave lens a convergent beam of rays becomes less convergent i.e., the convergent beam meets further away from the lens. In case of concave lens Q’ is more distant from the lens than Q. Obviously in this case image distance, OQ’ is greater than object distance, OQ [Fig.(b)].
But if the virtual object distance, OQ is greater than the focal length of the concave lens, the image formed by the concave lens becomes virtual.
Remember that virtual object distance is positive.
Numerical Examples
Example 1.
If a convex lens of focal length 20 cm is placed in the path of a convergent beam of rays, the beam meets at Q. In absence of the lens the beam would meet at P. If the distance of P from the lens is 30 cm, determine the distance of Q from the lens.
Solution:
The converging beam of rays meets at Q after refraction in the lens LL’. In absence of the lens the beam would meet at P [Fig.], In this case with respect to the lens, P is the virtual object and Q is its real image.
Here, u = 30 cm, f = 20 cm, v = OQ = ?
The equation of lens is \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{v}-\frac{1}{30}\) = \(\frac{1}{20}\) or, \(\frac{1}{v}\) = \(\frac{1}{30}+\frac{1}{20}\) = \(\frac{1}{12}\) or, v = 12 cm
∴ Required distance, OQ = 12 cm
Example 2.
A converging beam of rays after refraction in a concave lens of focal length 20 cm meets at a distance of 15 cm from the lens. In absence of the lens where would the
beam meet?
Solution:
In absence of the lens the converging beam would meet at P [Fig.]. So, with respect to this lens P is the virtual object and Q is its real image.
In this case, u = OQ = 15 cm; f = -20 cm
The equation of lens is \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{15}\) – \(\frac{1}{u}\) = –\(\frac{1}{20}\) or, \(\frac{1}{u}\) = \(\frac{1}{20}\) + \(\frac{1}{15}\)
∴ u = \(\frac{60}{7}\) = 8.57 cm
Therefore, in absence of the lens the beam of rays would meet at a distance of 8.57 cm from the lens.