In the given figure the radius of curvature of the curved face in the plano-convex and the plano-concave lens is $15\ \mathrm{cm}$ each, and the two lenses (kept $20\ \mathrm{cm}$ apart) are made of material of refractive index $1.5$. A parallel beam is incident on the combination; find the final position of the image formed.

The final image is real and is formed $15\ \mathrm{cm}$ beyond the plano-concave lens.

CBSE Class 12 · 2023 · Geometrical OpticsShort Answer

In the given figure the radius of curvature of the curved face in the plano-convex and the plano-concave lens is

15 cm

each, and the two lenses (kept

20 cm

apart) are made of material of refractive index

1.5

. A parallel beam is incident on the combination; find the final position of the image formed.

Answer: The final image is real and is formed $15 cm$ beyond the plano-concave lens.

\textbf{Plano-convex lens:}

\frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{\infty} - \frac{1}{- 15}) = \frac{0.5}{15} \Rightarrow f_{1} = + 30 cm

.
\textbf{Plano-concave lens:}

\frac{1}{f _{2}} = (1.5 - 1) (- \frac{1}{15} - \frac{1}{\infty}) \Rightarrow f_{2} = - 30 cm

.
The parallel beam converges at the focus of the first lens,

30 cm

beyond it. As the second lens is

20 cm

away, this convergence point lies

30 - 20 = 10 cm

\emph{past} it and acts as a virtual object,

u = + 10 cm

\frac{1}{v} - \frac{1}{u} = \frac{1}{f _{2}} \Rightarrow \frac{1}{v} = \frac{1}{f _{2}} + \frac{1}{u} = - \frac{1}{30} + \frac{1}{10} = \frac{- 1 + 3}{30} = \frac{1}{15}

v = + 15 cm (real image, 15 cm beyond the plano-concave lens)

CBSE Class 12 Board · 2023 · Subjective