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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
Focal length, f = 20 cm (a) Using the lens formula, 1v =1f+1u∴ 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
∴
1v=1f+1u =-116+112 = -3+448 =148
⇒ 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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