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

Engineering

Physics

Curved Surface Refraction

Question

40.0 cm

13.3 cm

–280.0 cm

21.5 cm

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$\frac{1}{f_{1}} = (1 .2 - 1) (\frac{1}{\infty} - \frac{1}{- 14}) = \frac{0 .2}{14}$

$\frac{1}{f_{2}} = (1 .5 - 1) (\frac{1}{14} - \frac{1}{\infty}) = \frac{0 .5}{14}$

$\frac{1}{f_{eq}} = \frac{0 .7}{14} = \frac{1}{20}$

$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20}$

v = 40 cm