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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.
10 Questions 40 Marks 10 Mins
Concept:
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) Where V = Distance of image from the optical center, U = Distance of object from the optical center
Calculation:
\(⇒ \frac{1}{f} =\frac{1}{V} -\frac{1}{U}\) \(⇒\frac{1}{f} =\frac{1}{X-Y} -\frac{1}{-Y}=\frac{1}{X-Y} +\frac{1}{Y}\) ⇒ 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
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