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 $40cm,$ the image distance will be [image:9_image55.emf]

Question

Dr. Abubacker · Accepted Answer

(b) [image:9_image120.emf]
$\frac{1}{f_{1}}(\mu - 1)\left\lbrack \frac{1}{R_{1}} - \frac{1}{R_{2}} \right\rbrack$
$\frac{1}{f_{1}} = (1.5 - 1)\left\lbrack \frac{1}{14} - \frac{1}{\infty} \right\rbrack$
$\frac{1}{f_{1}} = \frac{0.5}{14}$
$\frac{1}{f_{2}} = (1.2 - 1)\left\lbrack \frac{1}{\infty} - \frac{1}{- 14} \right\rbrack$
$\frac{1}{f_{2}} = \frac{0.2}{14}$
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{0.5}{14} + \frac{0.2}{14}$
$\Rightarrow \frac{1}{f} = \frac{0.7}{14}$
$\frac{1}{v} = \frac{7}{140} - \frac{1}{40} = \frac{1}{20} - \frac{1}{40}$
$\Rightarrow \frac{1}{v} = \frac{2 - 1}{40}$
$\Rightarrow v = 40\ cm$

Explanation

Reviewed by Dr. Abubacker

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