How is a Real Image Formed by a Convex Lens?

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A Few Problems on Formation of Real Image by a Convex Lens

i) Prove that by keeping the object and the screen fixed, a con-vex lens can be placed in two such positions that in each position, a distinct image of the object is formed on the screen.

Suppose, PQ is an object and S₁S₂ is a screen [Fig.]. In between them a convex lens LL’ is placed. The lens forms a real image pq on the screen.
Let the distance between the object and the screen = D, object distance = u, image distance = v.
Here, D = u + v
For formation of real image by a convex lens the equation of the lens is \(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{f}\)

∴ \(\frac{1}{D-u}+\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{D}{(D-u) u}\) = \(\frac{1}{f}\) or, u² – Du + Df = 0 ….. (1)
On solving this equation, the following two values of u are obtained:

For the different values of (D² – 4Df) three cases may arise:

1) When D² > 4Df i.e., D > 4f, the values of u₁ and u₂ are real and different.
So, if the distance between the object and the screen is greater than 4 times the focal length of the lens, then for two different positions of the lens, two real images of the object are formed on the screen.

2) When D² = 4Df i.e., D = 4f, the values of u₁ and u₂ are real and equal (u₁ = u₂ = \(\frac{D}{2}\))
So, if the distance between the object and the screen is equal to 4 times the focal length of the lens, then for a single position of the lens, real images of the object are formed on the screen.
In this case the lens is to be placed at the middle of the- object and the screen, because object distance = u = u₁ = u₂ = \(\frac{D}{2}\) = 2f and image distance = v = D – u = 4f – 2f = 2f.

3) When D² < 4Df i.e., D < 4f, the values of u₁ and u₂ are imaginary. So in this case, wherever the lens is placed, no image is formed on the screen.

So it is clear from the above discussion that to obtain a real image on a screen with the help of a convex lens, the minimum distance between the object and the screen should be 4 times the focal length of the lens. If the distance between the object and the screen is greater than 4 times the focal length of the lens, two images will be obtained on the screen for two different positions of the lens.

ii) An object and a screen are placed at a fixed distance D apart from each other. There are two positions for a convex lens in between them for a sharp image to be cast on the screen. If x be the distance of separation between these two positions, show that the focal length of the lens is given by f = \(\frac{D^2-x^2}{4 D}\).

Suppose, OO₁ is the position of the object and SS₁ is the position of the screen [Fig.], L₁ and L₂ are two different positions of the lens. For these two positions of the lens, the image of the object is obtained on the screen.

iii) If two images of the object are obtained on the screen for two different positions of the lens, prove that u₁ = v₂ and v₁ = u₂.
Suppose, the distance between the object and the screen is D.
For the first position of the lens at L₁, object distance = u₁ and image distance = v₁.
For the second position of the lens at L₂ the corresponding values are u₂ and v₂ respectively.
∴ u₁ + v₁ = D and u₂ + v₂ = D
Taking, u₁ = \(\frac{D+\sqrt{D^2-4 D f}}{2}\)
and so, u₂ = \(\frac{D-\sqrt{D^2-4 D f}}{2}\)
∴ v₁ = D – \(\frac{D+\sqrt{D^2-4 D f}}{2}\) = \(\frac{D-\sqrt{D^2-4 D f}}{2}\) = u₂
and v₂ = D – \(\frac{D-\sqrt{D^2-4 D f}}{2}\) = \(\frac{D+\sqrt{D^2-4 D f}}{2}\) = u₁

iv) A convex lens is placed between an object and a screen. If d₁ and d₂ be the lengths of the two real images formed for two positions of the lens and d be the length of the object, prove that d = \(\sqrt{d_1 d_2}\).
Suppose, for the first position of the lens the length of the image = d₁ and for the second position of the lens the length of the image = d₂.
If u₁ and v₁ be the object distance and the image distance respectively for the first position of the lens, then
magnification, m₁ = \(\frac{d_1}{d}\) = \(\frac{v_1}{u_1}\) = \(\frac{D-u_1}{u_1}\) [∵ D = u₁ + v₁]
If u₂ and v₂ be the object distance and the image distance for the second position of the lens, then

i.e., the size or length of the object are the geometrical mean of the sizes or lengths of the images formed on screen for two different positions of the lens.

v) A convex lens is placed between an object and a screen. A real image of the object is formed for two positions of the lens. If m₁ and m₂ be the magnifications of the image for the two positions of the lens respectively, prove that the focal length of the lens is given by f = \(\frac{x}{m_2-m_1}\) ; where x = distance between the two positions of the lens.

According to the Fig., for the position of the lens at L₁, object distance = u₁, and image distance = v₁. For the formation of real image by the convex lens the equation of the lens is
\(\frac{1}{v_1}+\frac{1}{u_1}\) = \(\frac{1}{f}\) or, 1 + \(\frac{v_1}{u_1}\) = \(\frac{v_1}{f}\)
or, 1 + m₁ = \(\frac{v_1}{f}\) [∵ m₁ = \(\frac{v_1}{u_1}\) ….. (3)
Similarly, for the position of the lens at L₂, object distance = u₂ and image distance = v₂.
For the formation of real image by the lens, the equation of the lens is

Displacement Method to find the Focal Length of A Convex Lens by Finding Position of Images

Procedure: Two pins N₁ and N₂ are mounted on the optical bench such that the separation between them is greater than four times the focal length of the convex lens. Now the lens is placed in between the pins N₁ and N₂. Two positions of the lens are found out in such way that in each position of the lens the image of one pin is formed at the position occupied by the other [Fig.].

Calculation: Let distance between the two pins be D and distance between the two positions of the lens be x.
For the first position L₁ of the lens,
u = N₁L₁ = \(\frac{1}{2}\)(N₁N₂ – L₁L₂) = \(\frac{1}{2}\)(D – x)
v = L₁N₂ = N₁N₂ – N₁L₁ = D – \(\frac{1}{2}\)(D – x) = \(\frac{1}{2}\)(D + x)
The general formula of a convex lens forming a real image is
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{2}{D+x}+\frac{2}{D-x}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{4 D}{D^2-x^2}\)
or, f = \(\frac{D^2-x^2}{4 D}\) ….. (1)
Measuring D and x from the optical bench, the focal length of a convex lens can be determined from equation (1).