A double convex lens made of glass of refractive index $1.6$ has both surfaces of equal radii of curvature of $30\ \mathrm{cm}$ each. An object of $5\ \mathrm{cm}$ height is placed at a distance of $12.5\ \mathrm{cm}$ from the lens. Find the position, nature and size of the image.

$v=-25\ \mathrm{cm}$ (virtual, erect, on the same side as the object); magnification $m=+2$; image size $=10\ \mathrm{cm}$, erect and magnified.

CBSE Class 12 · 2005 · All India · Geometrical OpticsShort Answer

A double convex lens made of glass of refractive index

1.6

has both surfaces of equal radii of curvature of

30 cm

each. An object of

5 cm

height is placed at a distance of

12.5 cm

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

Focal length:

\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.6 - 1) (\frac{1}{30} + \frac{1}{30}) = 0.6 \times \frac{2}{30} = \frac{1}{25} \Rightarrow f = + 25 cm

.
Object within the focal length (

u = - 12.5 cm < f

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{25} - \frac{1}{12.5} = \frac{1 - 2}{25} = - \frac{1}{25} \Rightarrow v = - 25 cm

m = \frac{v}{u} = \frac{- 25}{- 12.5} = + 2, h^{'} = m h = 2 \times 5 = 10 cm (virtual, erect, magnified)

CBSE Class 12 Board · 2005 · All India · Subjective