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The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (μ = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

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#### Solution

(a) Let *F* be the the focal length of the given concavo-convex lens. Then,

\[\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_m} + \frac{1}{f_1}\]

\[= \frac{2}{f_1} + \frac{1}{f_m}\]

\[\left[ \because \frac{1}{f_m} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \right]\]

\[\frac{1}{F} = \frac{2}{30} + \frac{1}{15} = \frac{2}{15}\]

Hence, *R* = 15 cm

Therefore, the pin should be placed at a distance of 15 cm from the lens.

(b) If the concave part is filled with water,

For focal length *F*'

\[\frac{1}{F'} = \frac{2}{f_w} + \frac{2}{f_1} + \frac{1}{f_m}\]

\[= \frac{2}{90} + \frac{2}{30} + \frac{1}{15}\]

\[\left[ \because \frac{1}{f_w} = \left( \frac{4}{3} - 1 \right)\left( + \frac{1}{30} \right) \right]\]

\[\therefore F' = \frac{45}{7} cm\]

Thus, pin should be placed at a distance of \[\frac{90}{7}\] cm from the lens.

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