Question
A thin equi-convex lens of flint glass (refractive index μ1) is kept coaxially in contact with another thin equi-concave lens of crown glass (refractive index μ2). The system is completely immersed in water (aμW = 4/3).
Parallel rays of light incident parallel to the principal axis in water are focused by this system at a distance of 24 cm beyond the system. The thickness of the system is negligible. If the radius of curvature of each surface is R = 20 cm, the difference (μ1 − μ2) is :
Options
Solution
[Key Concept: For a thin lens in a medium, 1/f = (μlens/μmedium − 1)(1/R1 − 1/R2); powers of lenses in contact add.]
For equiconvex lens
$$\frac{1}{f_{1}} = \left( \frac{\mu_{1}}{\mu_{\omega}} - 1 \right)\left( \frac{2}{R} \right)$$
For equiconcave lens
$$\frac{1}{f_{2}} = \left( \frac{\mu_{2}}{\mu_{\omega}} - 1 \right)\left( - \frac{2}{R} \right)$$
Combined focal length
$$\begin{matrix} & \frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ & \ = \frac{2}{R}\left\lbrack \left( \frac{\mu_{1}}{\mu_{w}} - 1 \right) - \left( \frac{\mu_{2}}{\mu_{\omega}} - 1 \right) \right\rbrack \\ & \frac{1}{f_{eq}} = \frac{2}{R\mu_{\omega}}\left( \mu_{1} - \mu_{2} \right) \\ & \frac{1}{24} = \frac{2}{20 \times \left( \frac{4}{3} \right)}\left( \mu_{1} - \mu_{2} \right) \\ & \mu_{1} - \mu_{2} = \frac{80}{144} = \frac{5}{9} \end{matrix} Answer\ is\ option\ d$$
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