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What do you Mean by Power of a Lens? What is its Unit?
The function of a lens is to converge a beam of light in case of a convex lens or to diverge it in case of a concave lens.
Definition: Power of a lens is the degree of convergence in case of convex lens or the degree of divergence in case of concave lens.
Since a convex lens of shorter focal length produces a greater convergence and a concave lens of shorter focal length produces a greater divergence, power of a lens varies inversely as its focal length i.e., power of a lens with shorter focal length is greater and power of a lens with larger focal length is smaller.
Hence, the reciprocal of the focal length of a lens is called its power. Representing focal length of a lens by f and power by P we can write, P = \(\frac{1}{f}\).
Unit of power of lens: Unit of power of lens is dioptre (dpt or, D).
1 dioptre is defined as the power of a lens of focal length one metre. So unit of power of lens in SI is m-1.
i.e., P \(=\frac{1}{\text { focal length in meter }}\) dioptre or m-1
\(=\frac{100}{\text { focal length in centimeter }}\) dipotre or m-1
or, P(dioptre) = \(\frac{1}{f(\text { metre })}\) = \(\frac{100}{f(\text { centimetre) }}\)
Power of a convex lens is considered positive and that of a concave lens is considered negative.
The convex lens having focal length 20 cm has power
= \(\frac{100}{20}\) = 5 dpt
The lens having power 4 dpt has focal length
= \(\frac{1}{4}\)m = 25 cm
Again, the nature of the lens having power -4 dioptre is concave and its focal length
f = –\(\frac{1}{4}\)m = -0.25m
Power of a combination of lenses: The power of a combination of lenses is equal to the algebraic sum of the powers of the constituent lenses.
The power of the combination of two lenses with powers P1 and P2 kept in contact with one another is
P = P1 + P2
Relation of radius of curvature of a lens with its power: We know that the equation of focal length of a lens is given by,
\(\frac{1}{f}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)
If the radii of curvature of the two surfaces of the lens be equal then in case of biconvex and biconcave lenses we have,
\(\frac{1}{f}\) = (µ – 1) ᐧ \(\frac{2}{r}\) and \(\frac{1}{f}\) = -(µ – 1) ᐧ \(\frac{2}{r}\) respectively
So, the corresponding power of these lenses are,
P = (µ – 1) ᐧ \(\frac{2}{r}\) and P = -(µ – 1) ᐧ \(\frac{2}{r}\) i.e., P ∝ \(\frac{1}{r}\)
Therefore, if the radius of curvature of a biconvex or a biconcave lens increases, power of the lens decreases and if the radius of curvature decreases power of the lens increases.
1. When two thin lenses having equal and opposite focal length (one convex and one concave) are placed in contact with each other and if F be their equivalent focal length then,
\(\frac{1}{F}\) = \(\frac{1}{f}\) + \(\frac{1}{-f}\) = 0 or, F = i = \(\frac{1}{0}\) = ∞
So, Power, p = \(\frac{1}{F}\) = 0
This type of combination of lenses acts as a plane glass plate.
2. If the refractive index of the medium surrounding the lens is greater than that of the material of the lens, the convex lens shows diverging power and the concave lens shows converging power.
3. Suppose that the refractive index of the surrounding medium is µm and the refractive index of the material of the lens is µg. If µm. increases, the power of the lens decreases i.e., its focal length increases.
Change of the focal length and the power of a lens with the wavelength of the incident light: According to Scientist Cauchy, the relation of wavelength (λ) with refractive index (µ) is given by
µ = A + \(\frac{B}{\lambda^2}\) [where A and B are constants]
Now, if f be the focal length of a lens then,
\(\frac{1}{f}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)
So, for any particular lens,
f ∝ \(\frac{1}{(\mu-1)}\)
In comparison to any other wavelength of light, wavelength of red light is greater i.e., for red light, refractive index of the lens is minimum. So the focal length of a lens is greater for red light is comparison to other colours.
Power of a lens is given by
P = \(\frac{1}{f}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\) or, P ∝ (µ – 1)
So, power of a lens will be less in case of red light in comparison to any other colour.
f-number of lens: The focal length of a lens is generally expressed as a multiple of the diameter of its aperture. This multiple is called f-number of lens. If F be the focal length and D be the diameter of the aperture, Then D = \(\frac{F}{f}\) or, f = \(\frac{F}{D}\). For example, if the f-number of a lens is (f – 15), it means that the diameter of the aperture of the lens is \(\frac{1}{15}\) of its focal length.
Depth of field: In optics, depth of field (DOF) is the distance between the nearest and farthest objects in a scene that appear acceptably sharp in an image. The term mainly relates to film and photography. For a particular lens greater the value of f-number, greater is the depth of field.
Numerical Examples
Example 1.
The power of a lens of a pair of spectacles is -2.5 dioptre. What is the nature and focal length of the lens used? [HS 2000]
Solution:
W know that, P = +\(\frac{100}{f}\) dioptre [if f is expressed in cm]
∴ -2.5 = +\(\frac{100}{f}\) or, f = -40 cm
Since focal length is negative, the lens is concave and its focal length is 40 cm.
Example 2.
A lens of power 0.5 dioptre is placed in front of a lens of power 2.0 dioptre. Assume that the distance between the two lenses is zero,
(i) What will be the power of the lenses in dioptre ?
(ii) What will be their focal lengths?
Solution:
i) Power of the combination of lenses,
P = P1 + P2 [here, P1 = 0.5 dpt, P2 = 2 dpt]
= 0.5 + 2 = 2.5 dpt = 2.5 m-1
ii) BM Focal length of the lens of power 0.5 dioptre,
f1 = \(\frac{+100}{0.5}\) = 200 em (convex lens)
and focal length of the lens of power 2.0 dpt,
f2 = \(\frac{100}{2}\) = +50 cm (convex lens).
Example 3.
A convex lens of focal length 40 cm is placed in contact with a concave lens of focal length 25 cm. What is the power of the lens combination?
Solution:
Power of the convex lens, P1 = \(\frac{100}{f_1}\) = \(\frac{100}{40}\) = 2.5 m-1
Power of the concave lens, P2 = \(\frac{100}{f_2}\) = \(\frac{100}{-25}\) = -4.0 m-1
∴ Power of the combination of lenses,
P = P1 + P2 = 2.5 – 4.0 = -1.5 m-1
Example 4.
An object is placed at a distance of 12 cm from a lens and a virtual image, four times magnified is formed. Calculate the focal length of the lens. Draw the ray diagram for the formation of image. What is the power of the lens?
Solution:
Magnification, m = \(\frac{v}{u}\) = 4 or, v = 4u
From diagram u = OQ = -12 cm, v = OQ’ = -48 cm
From the lens equation,
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{-48}-\frac{1}{-12}\)
or, \(\frac{1}{f}\) = \(\frac{1}{12}-\frac{1}{48}\) or, f = 16
∴ Focal length of the lens is 16 cm.
The lens is convex i.e., f is positive.
Power of the lens, P = \(\frac{100}{f}\)D = \(\frac{100}{16}\)D = +6.25 D
Example 5.
A real image of an object is formed by a convex lens on a screen at a distance of 20 cm from the lens. When a concave lens is placed at a distance of 5 cm from the convex lens towards the screen the image is shifted through 10 cm. Calculate the focal length and the power of the concave lens. [HS ‘02]
Solution:
The role of the convex lens is just to form the first image on the screen.
For the concave lens this image acts as the virtual object and another real image is formed at a distance of 10 cm away from the virtual object [Fig.],
Let P be the position of the object. L1 and L2 are the convex and concave lens respectively separated by a distance O1O2 = 5 cm
P’ is the real image formed by L1 and O1P’ = 20 cm
P’ acts as the virtual object for the second lens L2 and a real image of P’ is formed at P1, the distance P’P1 = 10 cm
∴ For the concave lens :
u = O2P’ = O1P’ – O1O2 = 20 – 5 = 15 cm and v = O2P1 = O2P’ + P’P1 = 15 + 10 = 25 cm
Putting in \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\), we get
\(\frac{1}{25}-\frac{1}{15}\) = \(\frac{1}{f}\) or, f = -37.5 cm
So, the focal length of the concave lens is 37.5 cm
Power of the lens = \(\frac{100}{f}\) = \(\frac{100}{-37.5}\) = -2.67D
Example 6.
The distance between a lamp and screen is 90 cm. Where should a convex lens of focal length 20 cm be placed in between the lamp and the screen so that a real image of the lamp is formed on the screen? [HS ‘03]
Solution:
Let the distance of the lamp from the lens be x cm.
So, image distance = (90 – x) cm i.e., u = -x and v = (90 – x).
For the formation of a real Image by the convex lens we have,
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{90-x}+\frac{1}{x}\) = \(\frac{1}{20}\) or, \(\frac{90}{x(90-x)}\) = \(\frac{1}{20}\)
or, x2 – 90x + 1800 = 0 or, (x – 30)(x – 60) = 0
or, x = 30 cm or, 60 cm
So, the lens is to be placed at a distance of 30 cm or 60 cm with respect to the lamp.
Example 7.
The size of the image of an object which is at infinity, as formed by a convex lens of focal length 30 cm, is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the Image at a distance of 26 cm from the convex lens, what will be the size of the final image? [IIT ‘03]
Solution:
For the convex lens, since the object is at infinity, image is formed at the focus (P’) [Fig.].
∴ v = f = 30 cm
This is the virtual object for the second lens which is at a distance O1O2 = 26 cm from the first lens.
∴ In this case object distance,
u = (O1P’ – O1O2) = 30 – 26 = 4 cm and f = -20 cm
∴ \(\frac{1}{v}-\frac{1}{4}\) = \(\frac{1}{-20}\) or, v = 5 cm
Magnification b;the second lens = \(\frac{5}{4}\)
∴ Size of the final image = size of the 1st image × \(\frac{5}{4}\)
= 2 × \(\frac{5}{4}\) = 2.5 cm
Example 8.
The image of an illuminated pin placed at a distance of 20 cm from a convex lens Is formed on a screen placed at a distance of 30 cm from the lens. In between the screen and the convex lens a concave lens is placed at a distance of 10 cm from the convex lens. If the screen is shifted through a distance of 10 cm more, a distinct image is formed on the screen. Find the ratio of the power of the two lenses. If the power of the concave tens is -4 m-1, how far should the screen Is to be shifted?
Solution:
For the convex 1ens
u = O1P = -20 cm, v = O1P = 20 cm, f = ?
From \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) we get
\(\frac{1}{30}-\frac{2}{-20}\) = \(\frac{1}{f_1}\) [let the focal length be f1]
∴ f1 = 12 cm
The image formed by the 1st lens is the virtual object of the second lens.
∴ For the second lens,
u = O2P = O1P – O1O2 = 30 – 10 = 20 cm
v = O2P’ = O2P – PP’ = 20 + 10 = 30 cm
f = f2 (say)
∴ From lens equation,
\(\frac{1}{30}-\frac{1}{20}\) = \(\frac{1}{f_2}\) or, f2 = -60 cm
Power of the convex lens, P1 = \(\frac{100}{f_1}\)
Power of the concave lens, P2 = \(\frac{100}{f_2}\)
∴ P1 : P2 = |f2| : |f1| = 60 : 12 = 5 : 1
If the power of the concave lens is -4 m-1, its focal length,
f = \(\frac{100}{P}\) = \(\frac{100}{-4}\) = -25 cm
Then u = 20 cm, f = -25 cm, v = ?
We know, \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{v}\) = \(\frac{1}{u}+\frac{1}{f}\) = \(\frac{1}{20}-\frac{1}{25}\) or, v = 1oo cm
∴ The screen has to be shifted by a distance (100 – 20) = 80 cm
Example 9.
The radii of curvature of the two surfaces of a convexo-concave lens made of glass are 20 cm and 60 cm. An object is situated on the left side of the lens at a distance of 80 cm along the principal axis.
(i) Determine the position of the image.
(ii) A similar type of lens is placed at a distance of 160 cm on the right side of the first lens co-axially. Determine the position of the image. [Given, µ of glass = 1.5]
Solution:
Focal length of a lens is given by lens maker’s formula,
\(\frac{1}{f}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)
in this case, µ = 1.5; r1 = +20 cm; r2 = +60 cm
∴ \(\frac{1}{f}\) = (1.5 – 1)\(\left(\frac{1}{20}-\frac{1}{60}\right)\)
or, f = 60 cm
i) Here, u = -80 cm, f = 60 cm
From \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) we get,
\(\frac{1}{v}+\frac{1}{80}\) = \(\frac{1}{60}\) or, v = 240 cm
So, image is at a distance 240 cm on the other side of the lens.
ii) The image P formed by the 1st lens will act as the virtual image of the second lens [Fig.].
∴ Object distance, u’ = O2P = 240 – 160 = 80 cm
[distance is measured along the direction of ray]
and f’ = 60 cm
∴ From \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\),
we get, \(\frac{1}{v}-\frac{1}{80}\) = \(\frac{1}{60}\) or, v = \(\frac{240}{7}\) = 34.28 cm
So, the final image is formed at a distance 34.28 cm from the second lens and on the other side of the first lens.
Example 10.
An illuminated object is placed at a distance 15 cm in front of a convex lens of focal length 12cm. If a plain mirror is placed behind the lens at a distance 45 cm, a distinct image is formed in front of the lens (in the object side) on the screen. Find the distance of the screen from the lens.
Solution:
LL’ is a convex lens and MM’ is a plain mirror placed at a distance 45 cm from the lens. P is an illuminated object at a distance 15 cm in front of the convex lens.
In case of image formation by the lens,
u = -15 cm, f = 12 cm, v = ?
∴ From lens equation,
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{v}+\frac{1}{15}\) = \(\frac{1}{12}\) or, \(\frac{1}{v}\) = \(\frac{1}{12}-\frac{1}{15}\) = \(\frac{1}{60}\)
∴ v = 60 cm
If plain mirror is absent a real image will be formed at P’. This P’ will be acted as virtual object for the plain mirror.
For plain mirror,
object distance = O1P’ = 60 – 45 = 15 cm
Mirror MM’ will form real image of P’ at P”.
Now, image P” acts as an object for formation of image in second time.
So, object distance = OP” = OO1 – O1P”
= (45 – 15) = 30 cm
[∵ O1P’ = O1P” = 15 cm]
Thus, u = -30 cm, f = 12 cm, v = ?
From lens equation, \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{1}{v}+\frac{1}{30}\) = \(\frac{1}{12}\) or, \(\frac{1}{v}\) = \(\frac{1}{12}-\frac{1}{30}\) = \(\frac{1}{20}\)
or, v = 20 cm
Therefore an image will be formed on the screen S at a distance 20 cm in front of the lens (in the object side).