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Some of the most important Physics Topics include energy, motion, and force.
What is an Equivalent Lens? What is the Formula of Equivalent Focal Length?
Equivalent lens: Suppose, the image of an object is produced by the combination of more than one co-axial lens. Now, without changing the position of the object and the image, a single lens is used in place of the combination. If this single lens produces the image of same magnification of the object and in the same position, this single lens is called the equivalent lens of the combination. The focal length of this lens is called equivalent focal length.
Equivalent focal length of the combination of two thin co-axial lenses in contact: Let two thin lenses L1 and L2 having focal lengths f1 and f2 respectively be placed in contact so as to have a common axis [Fig.], Since the lenses are thin, we may assume that the optical centres of the two lenses coincide at a single point. Suppose, the point O is their common optical centre.
P is a point-object on the principal axis. Rays of light starting from P form an image at the point Q1 due to refraction in the lens L1.
In this case, object distance = OP = -u, image distance = OQ1 = v1, focal length = +f1
distance = OQ = +v1, focal length = +f1
∴ \(\frac{1}{v_1}-\frac{1}{u}\) = \(\frac{1}{f_1}\) or \(\frac{1}{v_1}+\frac{1}{u}\) = \(\frac{1}{f_1}\) …. (1)
Q1 acts as a virtual object with respect to the second lens and forms the final real image at Q due to refraction in the second lens L2. So, in this case, object distance = OQ1 = + v1, image distance = QQ = +v, focal length = f2.
∴ \(\frac{1}{v}-\frac{1}{v_1}\) = \(\frac{1}{f_2}\) …. (2)
Adding equations (1) and (2) we get,
\(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{f_1}+\frac{1}{f_2}\) …. (3)
Now, if equivalent focal Length be F, according to definition, the equivalent lens will form the image of the object P at Q i.e., the equivalent lens will be converging. So, in case of equivalent lens we can write,
\(\frac{1}{v}-\frac{1}{-u}\) = \(\frac{1}{F}\) or, \(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{F}\) …. (4)
From equations (3) and (4), we get
\(\frac{1}{F}\) = \(\frac{1}{f_1}+\frac{1}{f_2}\) ….. (5)
If a number of thin lenses are placed in contact, it can be proved similarly that
\(\frac{1}{F}\) = \(\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}\) + ……. (6)
It is to be noted that instead of the combination of convex lenses, a combination of concave Lenses or mixed combination of con-vex and concave lenses may be used. In each case, equivalent focal length is obtained from equation (6), with proper signs of focal lengths of the lenses used.
If an equi-convex lens of focal length f is cut vertically into two equal haves, each half wifi have the focal length equal to 2f.
One of the lenses of the combination is convex and the other is concave and the other is concave : Let the focal length of the convex lens be f1 and that of the concave lens be f2. If F be the focal length of the combination, then
\(\frac{1}{F}\) = \(\frac{1}{f_1}+\frac{1}{-f_2}\) = \(\frac{1}{f_1}-\frac{1}{f_2}\) ∴ F = \(\frac{f_1 f_2}{f_2-f_1}\)
- If f1 > f2 then F vill be negative and the combination will act as a concave lens.
- If f1 < f2 then F will be positive and the combination will act as a convex lens.
- If f1 = f2, then F will be infinite and the combination will act as a plane laminar plate.
General formula of equivalent focal length: In all practical purposes no lens-combination is formed by keeping the lenses in contact. Rather, in different optical instruments the lenses are necessarily placed separated by a distance. Again, even when the two lenses are placed in contact, an effective distance due to their thickness is introduced. This distance cannot be ignored in all cases.
Let the focal length of two lenses placed co-axially be f1 and f2 and the distance between their optical centres be a.
Then the expression of equivalent focal length of the combination of lenses is given by,
\(\frac{1}{F}\) = \(\frac{1}{f_1}+\frac{1}{f_2}-\frac{a}{f_1 f_2}\) ….. (7)
When the two lenses are in contact, a = 0; then equation (7) becomes equation (5).
Numerical Examples
Example 1.
A combination of a convex lens of focal length 20 cm and a concave lens of focal length 10 cm is formed by keeping in contact with each other. Determine the equivalent focal length of the combination.
Solution:
Here, f1 = 20 cm and f2 = -10 cm
We know, \(\frac{1}{F}\) = \(\frac{1}{f_1}+\frac{1}{f_2}\)
∴ \(\frac{1}{F}\) = \(\frac{1}{20}+\frac{1}{-10}\) = \(\frac{-1}{20}\) or, F = -20 cm
Since the sign of equivalent focal length is negative, the equivalent lens is a concave lens.
Example 2.
A lens-combination is formed by keeping a lens in con-tact with a concave lens of focal length 25 cm. This lens- combination produces a real image magnified 5 times, of an object placed at a distance of 20 cm from the combination. Calculate the focal length and the nature of the lens placed in contact with the concave lens.
Solution:
Final magnification, m = \(\frac{v}{u}\) = 5
∴ v = 5u = 100 cm (as u = 20 cm)
From lens formula, = \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{F}\)
we get, for the combination
\(\frac{1}{100}-\frac{1}{-20}\) = \(\frac{1}{F}\) [ ∵ u is negative]
or, F = \(\frac{100}{6}\) = 16.66 cm
For the combination, let focal length of the unknown lens be f. Then
\(\frac{1}{f}-\frac{1}{25}\) = \(\frac{1}{16.66}\) or \(\frac{1}{f}\) = \(\frac{1}{16.66}+\frac{1}{25}\)
or, f = 10 cm (approx.)
∴ The unknown lens is a convex lens of focal length 10 cm.
Combination Of Lenses And Mirror
Combination of a convex lens and a plane mirror: Determination of the focal length of a convex lens: A plane mirror MM’ is placed behind a convex lens LL’.
[Fig.]. A pin AB is placed in front of the combination of lens and mirror in such a way that the tip of the pin A just touches the principal axis of the lens. The pin is moved until its real image A’B’ coincides with the pin itself without parallax. This is possible only if the rays from A are incident normally on the plane mirror after refraction from the lens and these rays retrace the same path.
Then the real image A’B’ is formed at the position of AB. The tip A of the pin AB indicates the position of focus of the lens. So if the lens is thin, the distance of the tip of the pin from the surface of the lens is the focal length of the lens. If the lens is thick, half of the thickness of the lens is to be added to the previous distance to get the focal length of the convex lens.
Combination of a convex lens and a convex mirror: Determination of the focal length of a convex mirror: In the Fig. LL’ is a convex lens. The convex mirror MM’ is placed a little behind the convex lens. Now the point object P is to be placed in front of the combination of lens and mirror at such a distance that the image of the object coincides with the point object. It is possible only if the rays from P are incident normally on the convex mirror after refraction by the lens. These rays if produced further, must converge to the centre of curvature C of the mirror. Hence the rays after being reflected from the convex mirror return along the same path and form the image at P.
Now on applying the lens formula for formation of a real image by a convex lens, the image distance OC is determined. Measuring the distance of the mirror from the lens i.e., OO1, the distance O1C is determined. This distance O1C is the radius of curvature of the convex mirror.
Half of this distance is the focal length of the convex mirror.
Combination of a convex lens and a concave mirror: Determination of focal length of a concave mirror: In the Fig. LL is a convex lens. P is a point object placed on the principal axis of the lens. MM’ is a concave mirror placed at a certain distance on the other side of the lens. The point object P is placed on the axis at such a distance that its image coincides with the object at the same point P.
Since the final image coincides with the object at the same point, it can be said that the ray of light after passing through the convex lens is incident on the mirror perpendicularly. So the point Q in the figure is the centre of curvature of the concave mirror. On applying the lens formula for formation of a real image by a convex lens the image distance OQ is determined.
So the radius of curvature of the concave mirror, QO1 = OO1-OQ
If OQ1 is known, QO1 can be determined.
Therefore focal length of the concave mirror = \(\frac{1}{2}\)QQ1
Combination of a convcave lens and a concave mirror: Determination of focal length of a concave lens: In the Fig. LL’ is a concave lens. P is a pin. It is taken as a point object placed on the principal axis of the lens. MM’ is concave mirror placed at a certain distance on the other side of the lens.
Now the pin is placed in front of the lens at such a distance that the image formed by the combination of the lens and the mirror will be formed at the position of the pin. It is possible only if the light rays from the object after refraction by the lens are incident perpendicularly on the concave mirror and after reflection from the mirror return along the same path. In that case if the lens is absent, the reflected rays would meet at Q, the centre of curvature of the mirror. So with respect to the concave mirror the real image of the virtual object at Q is formed at P.
If the radius of curvature of the concave mirror is known, QO’ will be known. Again if the distance between the lens and the mirror i.e., OO’ is known, QO will be known since QO = QO’ – OO’.
This QO is the virtual object distance with respect to the concave mirror. Knowing the distance PO and using the general lens for-mula, the focal length of the concave lens can be determined.