Question

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.
Figure

Answer 1

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

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_m}+\frac{1}{f_1}$
$=\frac{2}{f_1}+\frac{1}{f_m}$ $\left[\because \frac{1}{f_m}=(\mathrm{\mu }-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\right]$



$\frac{1}{F}=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}$
⇒ F = 7.5 cm

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'
​ ​$\frac{1}{F\mathit{\text{'}}}=\frac{2}{f_w}+\frac{2}{f_1}+\frac{1}{f_m}$
$=\frac{2}{90}+\frac{2}{30}+\frac{1}{15}$ $\left[\because \frac{1}{f_w}=\left(\frac{4}{3}-1\right)\left(+\frac{1}{30}\right)\right]$
$\therefore F\text{'}=\frac{45}{7}\mathrm{cm}$
Thus, pin should be placed at a distance of $\frac{90}{7}$ cm from the lens.