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Ray Optics And Optical Instruments

Question
CBSEENPH12047830

For a given lens, the magnification was found to be
twice as large as when the object was 0.15 m
distant from it as when the distance was 0.2 m. The
focal length of the lens is

  • 1.5 m

  • 0.20 m

  • 0.10 m

  • 0.05 m

Solution

B.

0.20 m

Let as shown, 1 and 2 are positions of objects and
images in two different situations.

It is given

   

               v1u1 = 2 v2u2

Here, u1 = -15 cm, u2= -20 cm

                  v1 = 2 v2 × u1u2

                      = 2 v2 × 1520

                   v1 = 32 v2

Now,           1f = 1v - 1u

                  1f = 1v1 - 1u1

                  1f = 1v2- 1u2

So,        1v1 - 1u1 = 1v2 - 1u2

⇒           23v2 + 115 = 1v2+120

⇒                        v = 20 cm

⇒                         v = 0.20 m