A thin convex lens L (refractive index = 1.5) | vedclass

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(B) For the image to form at the object itself,the rays must retrace their path back to the object. This implies that the rays must be incident on the

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$\frac{4}{3}$

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(B) For the image to form at the object itself,the rays must retrace their path back to the object. This implies that the rays must be incident on the plane mirror $M$ normally.Case $1$: When the lens is directly on the mirror,the object at $A$ must be at the focal point of the lens. Given $OA = f = 18\, cm$.Using the lens maker's formula for a symmetric convex lens $(R_1 = R, R_2 = -R)$:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (1.5 - 1) \left( \frac{2}{R} \right) = \frac{1}{R}$Since $f = 18\, cm$,we have $R = 18\, cm$.Case $2$: $A$ liquid lens is formed between the convex lens and the mirror. This liquid lens is a plano-concave lens with radius of curvature $R = 18\, cm$ for the curved surface and $\infty$ for the flat surface.The focal length of the liquid lens $f_l$ is given by:$\frac{1}{f_l} = (\mu_l - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{(\mu_l - 1)}{R}$The combination of the convex lens $(f_1 = 18\, cm)$ and the liquid lens $(f_l)$ acts as a single lens with focal length $F = OA' = 27\, cm$.Using the combination formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_l}$:$\frac{1}{27} = \frac{1}{18} - \frac{(\mu_l - 1)}{18}$$\frac{1}{27} = \frac{1 - \mu_l + 1}{18} = \frac{2 - \mu_l}{18}$$18 = 27(2 - \mu_l)$$2 = 3(2 - \mu_l)$$2 = 6 - 3\mu_l$$3\mu_l = 4$$\mu_l = \frac{4}{3}$

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