(a) Under what condition will a glass lens placed in a transparent liquid become invisible ? (b) Describe and illustrate with a diagram, how we should arrange two converging lenses so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens. (c) An object is placed at a distance of $3\text{ cm}$ from a concave lens of focal length $12\text{ cm}$. Find the (i) position and (ii) nature of the image formed. OR (a) With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light. (b) A $2.0\text{ cm}$ tall object is placed perpendicular to the principal axis of a concave lens of focal length $15\text{ cm}$. At what distance from the lens, should the object be placed so that it forms an image $10\text{ cm}$ from the lens ? Also find the nature and the size of image formed.

(a) A glass lens becomes invisible when placed in a transparent liquid if the refractive index of the glass is exactly equal to the refractive index of the liquid. In this condition, no bending of light (refraction) occurs at the interface, and the lens does not reflect or refract light differently from the surrounding medium.

(b) To make a parallel beam emerge parallel after passing through two converging (convex) lenses, they must be placed coaxially such that the distance between them is equal to the sum of their focal lengths ($d = f_1 + f_2$). The principal focus of the first lens must coincide with the principal focus of the second lens.

(c) Given for concave lens: Focal length $f = -12\text{ cm}$, Object distance $u = -3\text{ cm}$.

(i) Using lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{-12} + \frac{1}{-3} = \frac{-1 - 4}{12} = \frac{-5}{12}$

$v = -2.4\text{ cm}$. The image is formed at $2.4\text{ cm}$ from the lens on the same side as the object.

(ii) Nature: Since $v$ is negative and it is a concave lens, the image is virtual, erect, and diminished.

OR

(a) A concave lens is thinner at the middle than at the edges. When a parallel beam of light falls on it, the rays undergo refraction at both surfaces. Due to the lens's shape, the rays bend away from the principal axis. When these refracted rays are produced backwards, they appear to diverge from a single point called the principal focus ($F$).

(b) Given: $h = 2.0\text{ cm}$, $f = -15\text{ cm}$, $v = -10\text{ cm}$ (Image is virtual for a concave lens).

Using lens formula: $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-10} - \frac{1}{-15} = \frac{-3 + 2}{30} = -\frac{1}{30}$

$u = -30\text{ cm}$. The object should be placed at $30\text{ cm}$ from the lens.

Nature: Virtual and erect.

Size: $m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3}$.

$h' = m \times h = \frac{1}{3} \times 2.0 = 0.67\text{ cm}$.