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 $(\mu =\frac{4}{3})$, find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

Answer 1

Let F is the focal length of given concavo-convex lens. Then,

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_m}+\frac{1}{f_1}$

$=\frac{2}{f_1}+\frac{1}{f_m}$

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

$=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}$

$\Rightarrow F=7.5\text{cm}$

Hence, R = 15 cm

Therefore, pin should be placed 15 cm from the lens.

(b) When water is filled in concave part.

For focal length F

$\frac{1}{F^{{}^{\prime }}}=\frac{2}{F_{iv}}+\frac{2}{F_1}+\frac{1}{F_m}$

$=\frac{2}{90}+\frac{2}{30}+\frac{1}{15}$

$[\because \frac{1}{F_w}=(\frac{4}{3}-1)(+\frac{1}{30})]$

$\therefore F^{{}^{\prime }}=\frac{45}{7}\text{cm}$

So, pin should be placed at $\frac{90}{7}\text{cm}$ from the lens.