List I contains four combinations of two lenses (1 and 2) whose focal lengths (in cm) are indicated in the figures. In all cases, the object is placed 20 cm from the first lens on the left, and the distance between the two lenses is 5 cm. List II contains the positions of the final images.

List-I	List-II
(I)	(P) Final image is formed at 7.5 cm on the right side of lens 2.
(II).	(Q) Final image is formed at 60.0 cm on the right side of lens 2.
(III)	(R) Final image is formed at 30.0 cm on the left side of lens 2.
(IV)	(S) Final image is formed at 6.0 cm on the right side of lens 2.
	(T) Final image is formed at 30.0 cm on the right side of lens 2.

Which one of the following options is correct?

The correct option is I → P, II → R, III → Q, IV → T.

To find the position of the final image formed by the system of two lenses, we apply the thin lens formula sequentially for lens 1 and lens 2.

The thin lens formula is given by:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

where:
• u = object distance from the lens
• v = image distance from the lens
• f = focal length of the lens

In all cases, the initial object O is placed at a distance of 20 cm to the left of Lens 1, so u₁ = -20 cm. The distance between Lens 1 and Lens 2 is d = 5 cm.

Case (I):

• Lens 1: Convex lens with focal length f₁ = +10 cm.
• Lens 2: Convex lens with focal length f₂ = +15 cm.

For Lens 1:

$\frac{1}{v_1}-\frac{1}{-20}=\frac{1}{10}$

$\frac{1}{v_1}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\Rightarrow v_1=+20\text{cm}$

The first image is formed 20 cm to the right of Lens 1. Since Lens 2 is located 5 cm to the right of Lens 1, this image acts as a virtual object for Lens 2 at a distance:

$u_2=+20-5=+15\text{cm}$

For Lens 2:

$\frac{1}{v_2}-\frac{1}{15}=\frac{1}{15}$

$\frac{1}{v_2}=\frac{2}{15}\Rightarrow v_2=+7.5\text{cm}$

Thus, the final image is formed at 7.5 cm on the right side of Lens 2. (I → P)

Case (II):

• Lens 1: Convex lens with focal length f₁ = +10 cm.
• Lens 2: Concave lens with focal length f₂ = -10 cm.

From Lens 1, the image is formed at v₁ = +20 cm to the right of Lens 1.
For Lens 2, the virtual object distance is u₂ = +15 cm.

For Lens 2:

$\frac{1}{v_2}-\frac{1}{15}=\frac{1}{-10}$

$\frac{1}{v_2}=-\frac{1}{10}+\frac{1}{15}=\frac{-3+2}{30}=-\frac{1}{30}\Rightarrow v_2=-30\text{cm}$

Thus, the final image is formed at 30.0 cm on the left side of Lens 2. (II → R)

Case (III):

• Lens 1: Convex lens with focal length f₁ = +10 cm.
• Lens 2: Concave lens with focal length f₂ = -20 cm.

From Lens 1, the image is formed at v₁ = +20 cm to the right of Lens 1.
For Lens 2, the virtual object distance is u₂ = +15 cm.

For Lens 2:

$\frac{1}{v_2}-\frac{1}{15}=\frac{1}{-20}$

$\frac{1}{v_2}=-\frac{1}{20}+\frac{1}{15}=\frac{-3+4}{60}=\frac{1}{60}\Rightarrow v_2=+60.0\text{cm}$

Thus, the final image is formed at 60.0 cm on the right side of Lens 2. (III → Q)

Case (IV):

• Lens 1: Concave lens with focal length f₁ = -20 cm.
• Lens 2: Convex lens with focal length f₂ = +10 cm.

For Lens 1:

$\frac{1}{v_1}-\frac{1}{-20}=\frac{1}{-20}$

$\frac{1}{v_1}=-\frac{1}{20}-\frac{1}{20}=-\frac{1}{10}\Rightarrow v_1=-10\text{cm}$

The image is formed 10 cm to the left of Lens 1. Since Lens 2 is 5 cm to the right of Lens 1, the object distance for Lens 2 is:

$u_2=-(10+5)=-15\text{cm}$

For Lens 2:

$\frac{1}{v_2}-\frac{1}{-15}=\frac{1}{10}$

$\frac{1}{v_2}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}\Rightarrow v_2=+30.0\text{cm}$

Thus, the final image is formed at 30.0 cm on the right side of Lens 2. (IV → T)

Combining all the matched pairs gives:
I → P, II → R, III → Q, IV → T