What is the minimum distance between an object and its real image formed by convex lens of focal length 20cm?

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A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of a focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Here, the point P on the right of the lens acts as a virtual object.

Object distance, u = 12 cm

∴           1v =120+112      =3+560 =860

i.e.,           v = 60/8 = 7.5 cm. 

Image is at a distance of 7.5 cm to the right of the lens, where the beam converges. (b)Now,Focal length of concave lens, f = –16 cmObject distance, u = 12 cm 

⇒      v = 48 cm 

Hence, the image is at a distance of 48 cm to the right of the lens, where the beam would converge.

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10 Questions 40 Marks 10 Mins

Concept:

• Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
  • Convex lens: ​A lens having two spherical surfaces, bulging outwards is called a double convex lens (or simply convex lens).
  • It is thicker in the middle as compared to the edges.
  • Convex lenses converge light rays and hence, convex lenses are also called converging lenses.

• The lens formula is given by:

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

Where V = Distance of image from the optical center, U = Distance of object from the optical center

• Depending upon the situations convex lens produces real and virtual images.

Calculation:

• Assume the distance between the object and the real image formed by the convex lens be' X'
• Let the distance of the object from the lens be, U = -Y
• So the image distance from the lens will be, V = X - Y
• The thin lens equation is given by

\(⇒ \frac{1}{f} =\frac{1}{V} -\frac{1}{U}\)

\(⇒\frac{1}{f} =\frac{1}{X-Y} -\frac{1}{-Y}=\frac{1}{X-Y} +\frac{1}{Y}\)

⇒ Y2 - XY + fX = 0

Then the value of Y according to the quadratic equation is given by, 

\(\Rightarrow Y = {-X \pm \sqrt{X^2-4fX} \over 2}\)

For the real value of Y, the value \(-X + \sqrt{X^2-4fX}≥ 0\)

⇒ X2 - 4fX ≥ 0

⇒ X2 ≥  4fx

⇒ X ≥ 4f 

• The minimum distance between the and the real image in a convex lens is 4f.

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