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?

Solution

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,

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

∴

\frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3 + 5}{60} = \frac{8}{60}

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 cm
Object distance, u = 12 cm

∴

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = - \frac{1}{16} + \frac{1}{12} = \frac{- 3 + 4}{48} = \frac{1}{48}

\Rightarrow

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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Wave Optics

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?

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