Question

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

Answer 1

In the given situation,the object is virtual and the image formed is real.Object distance,u=+12cm
(a)Focal length of the convex lens,f=20cm
Image distance\(\frac{1}{v}\)=v
According to the lens formula,we have the relation:

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

=\(\frac{1}{v}-\frac{1}{12}=\frac{1}{20}\)

=\(\frac{1}{v}=\frac{1}{20}+\frac{1}{12}\)

=3+\(\frac{5}{60}\)

=\(\frac{8}{60}\)
∴v=\(\frac{60}{8}\)=7.5cm
Hence, the image is formed 7.5cm away from the lens, towards its right.
(b)Focal length of the concave lens,f=-16cm
Image distance=v
According,to the lens formula, we have the relation: \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}\)=\(-\frac{1}{16}+\frac{1}{12}\)=\(-3+\frac{4}{48}\)=\(\frac{1}{48}\) ∴v=48cm
Hence, the image is formed 48cm away from the lens, towards its right.