A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index of the first lens is $1.5$ and that of second is $1.2$ . Both the curved surfaces are of the same radius of curvature $R = 14$ cm. For an object distance of $40$ cm, the image distance will be

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The given bi-convex lens can be considered as a combination of two plano-convex lenses placed in contact with their flat surfaces facing each other.

For the first plano-convex lens (left):
Refractive index, $n_{1} = 1.5$
Radius of curvature of the first surface, $R_{1} = + 14$ cm
Radius of curvature of the second surface, $R_{2} = \infty$
Using the lens maker's formula, its focal length $f_{1}$ is:
$\frac{1}{f _{1}} = (n_{1} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{14} - \frac{1}{\infty}) = \frac{0.5}{14} = \frac{1}{28} cm^{- 1}$

For the second plano-convex lens (right):
Refractive index, $n_{2} = 1.2$
Radius of curvature of the first surface, $R_{1} = \infty$
Radius of curvature of the second surface, $R_{2} = - 14$ cm
Its focal length $f_{2}$ is:
$\frac{1}{f _{2}} = (n_{2} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f _{2}} = (1.2 - 1) (\frac{1}{\infty} - \frac{1}{- 14}) = \frac{0.2}{14} = \frac{1}{70} cm^{- 1}$

The equivalent focal length $F$ of the combination is given by:
$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$
$\frac{1}{F} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20} cm^{- 1}$
$\Rightarrow F = + 20$ cm

Given the object distance $u = - 40$ cm, we use the lens formula to find the image distance $v$ :
$\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$
$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20}$
$\frac{1}{v} + \frac{1}{40} = \frac{1}{20}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40}$
$\Rightarrow v = + 40.0$ cm

Answer: $40.0$ cm

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