A double convex lens made of glass of refractive index $1.5$ has both surfaces of equal radii of curvature of $20\ \mathrm{cm}$ each. An object of $5\ \mathrm{cm}$ height is placed at a distance of $10\ \mathrm{cm}$ from the lens. Find the position, nature and size of the image.

$v=-20\ \mathrm{cm}$ (virtual, erect); magnification $m=+2$; image size $=10\ \mathrm{cm}$.

CBSE Class 12 · 2005 · Geometrical OpticsShort Answer

A double convex lens made of glass of refractive index

1.5

has both surfaces of equal radii of curvature of

20 cm

each. An object of

5 cm

height is placed at a distance of

10 cm

from the lens. Find the position, nature and size of the image.

Answer: $v = - 20 cm$ (virtual, erect); magnification $m = + 2$ ; image size $= 10 cm$ .

Focal length:

\frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) = 0.5 \times \frac{2}{20} = \frac{1}{20} \Rightarrow f = + 20 cm

.
With

u = - 10 cm

(inside the focus):

\frac{1}{v} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = - \frac{1}{20} \Rightarrow v = - 20 cm

m = \frac{v}{u} = \frac{- 20}{- 10} = + 2, h^{'} = 2 \times 5 = 10 cm (virtual, erect, magnified)

CBSE Class 12 Board · 2005 · Subjective